Prove:
$$
\begin{array}{|cccccccccc|} 1 & 0 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & 0 \\ x & x & x & \cdots & x & y & y & y & \cdots & y \\ x^{2} & 2 x^{2} & 2^{2} x^{2} & \cdots & 2^{m-1} x^{2} & y^{2} & 2 y^{2} & 2^{2} y^{2} & \cdots & 2^{m-1} y^{2} \\ x^{3} & 3 x^{3} & 3^{2} x^{3} & \cdots & 3^{m-1} x^{3} & y^{3} & 3 y^{3} & 3^{2} y^{3} & \cdots & 3^{m-1} y^{3} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ x^{n} & n x^{n} & n^{2} x^{n} & \cdots & n^{m-1} x^{n} & y^{n} & n y^{n} & n^{2} y^{n} & \cdots & n^{m-1} y^{n} \end{array}=(x-y)^{m^{2}}(x y)^{\frac{m^{2}-m}{2}}\left(\prod_{i=0}^{m-1} i !\right)^{2}
$$
where $n=2m-1$.
A friend of mine gave me this problem. He claimed that he solved this by a very complicated method(which is too long to type here, and to be frank, I didn't get it at all). The following part is my progress.
First, we want to extract $y$ so that we can let $z=\frac{x}{y}$, and then the determinant should be a polynomial $f(z)$ of variable $z$. Thus we can use calculus to simplify it. This trick usually works nicely on two-varible homogeneous determinant, but it doesn't kill this problem. Since we know the original problem is equivalent to
$$
\begin{array}{|cccccccccc|} 1 & 0 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & 0 \\ x & x & x & \cdots & x & 1 & 1 & 1 & \cdots & 1 \\ x^{2} & 2 x^{2} & 2^{2} x^{2} & \cdots & 2^{m-1} x^{2} & 1 & 2 & 2^{2} & \cdots & 2^{m-1} \\ x^{3} & 3 x^{3} & 3^{2} x^{3} & \cdots & 3^{m-1} x^{3} & 1 & 3 & 3^{2} & \cdots & 3^{m-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ x^{n} & n x^{n} & n^{2} x^{n} & \cdots & n^{m-1} x^{n} & 1 & n & n^{2} & \cdots & n^{m-1} \end{array}=(x-1)^{m^{2}}x^{\frac{m^{2}-m}{2}}\left(\prod_{i=0}^{m-1} i !\right)^{2}
$$
Note $f(x)=LHS$, calculate the derivative of $f(x)$, we have
$$
x\frac{\,\mathrm{d}}{\,\mathrm{d}x}f(x)=\begin{array}{|cccccccccc|}
1 & 0 & \cdots & 0 & 0 & 1 & 0 & 0 & \cdots & 0 \\
x & x & \cdots & x & x & 1 & 1 & 1 & \cdots & 1 \\
x^{2} & 2 x^{2} & \cdots & 2^{m-2} x^{2} & 2^{m} x^{2} & 1 & 2 & 2^{2} & \cdots & 2^{m-1} \\
x^{3} & 3 x^{3} & \cdots & 3^{m-2} x^{3} & 3^{m} x^{3} & 1 & 3 & 3^{2} & \cdots & 3^{m-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
x^{n} & n x^{n} & \cdots & n^{m-2} x^{n} & n^{m} x^{n} & 1 & n & n^{2} & \cdots & n^{m-1}
\end{array}
$$
which doesn't help at all.
I hope you can share some thoughts of this problem.
Best Answer
We will execute a "degree matching argument". That is,
The determinant of the matrix $$ \begin{array}{|cccccccccc|} 1 & 0 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & 0 \\ x & x & x & \cdots & x & 1 & 1 & 1 & \cdots & 1 \\ x^{2} & 2 x^{2} & 2^{2} x^{2} & \cdots & 2^{m-1} x^{2} & 1 & 2 & 2^{2} & \cdots & 2^{m-1} \\ x^{3} & 3 x^{3} & 3^{2} x^{3} & \cdots & 3^{m-1} x^{3} & 1 & 3 & 3^{2} & \cdots & 3^{m-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ x^{n} & n x^{n} & n^{2} x^{n} & \cdots & n^{m-1} x^{n} & 1 & n & n^{2} & \cdots & n^{m-1} \end{array} $$ is a polynomial in $x$ (by the Laplace expansion). Its degree can also clearly be upper bounded by the same expansion.
Then, we will show that $(x-1)^{m^2}$ and $x^{\frac{m^2-m}{2}}$ divide this determinant, as polynomials. Since these are co-prime polynomials, we will have then shown that the determinant is divisible by $(x-1)^{m^2}x^{\frac{m^2-m}{2}}$, and hence determined up to a constant.
This constant can then be found by matching the coefficient of the leading power of $x$ in this determinant to the coefficient of the leading power of $x$ in $(x-1)^{m^2}x^{\frac{m^2-m}{2}}$ (which is obviously $1$ by the binomial expansion).
Keep in mind that this argument is long in that writing down the technicalities rigorously is hard, but it's not difficult to come up with because it presents itself as a consequence of wanting to create the required divisors $x^{(m^2-m)/2}$ and $(x-1)^{m^2}$, as I explain later.
Step 1
To bound the degree of the determinant as a polynomial, observe that the largest power of $x$ that can be created is as follows : we match each of the last $m$ rows to one of the first $m$ columns bijectively and note that this bijection will lead to a term in the Laplace expansion by picking those entries as part of a term in the expansion.
For example, if $m=2$ then we'd have the term $x^2, 2x^3$ paired up, and $x^3,2x^2$ paired up. These lead to the largest degree, rather obviously.
That largest degree is seen to be $m + (m+1)+\ldots+(2m-1) = \frac{2m(2m-1)}{2}-\frac{m(m-1)}{2} = \frac{3m^2-m}{2} = m^2 + \frac{m^2-m}{2}$.
Note that this is only an upper bound on the degree because, for all we know, the coefficient of this monomial,which depends upon the rest of the matrix, could end up being $0$.
Step 2
We start by thinking about how $(x-1)^{m^2}$ can divide the determinant. I mean, the number of rows is itself $2m-1$, so how are we going to get that large a power of $(x-1)$ to divide the determinant?
The answer lies in creating linear combinations that result in large powers of $(x-1)^n$. To explain this, let us define the following transformation. Given vectors (or real numbers, if they are one-dimensional vectors) $v_1,v_2,\ldots,v_k$, define $$ T_k(v_1,v_2,\ldots,v_k) = v_1 - \binom{k}{k-1} v_2 + \binom{k}{k-2}v_3 - \ldots \pm v_k. $$
Keep in mind that, by the binomial expansion, $$ (x-1)^n = x^n - \binom{n}{n-1}x^{n-1} + \binom{n}{n-2}x^{n-2} - \ldots \pm 1 = T_n(x^n,x^{n-1},\ldots,1). $$
So we know that this kind of transformation will lead to something productive, given that powers of $x$ lie along the rows.
The next thing we want to do is to succinctly capture the way the entries are changing in each row. For this, define the transformation $$ M(p(x)) = x \frac{d}{dx}p(x). $$ for a polynomial $p(x)$. With this notation, see $M(x^l) = x \times lx^{l-1} = lx^l$. Then $M^2(x^n) = l^2x^l$, and so on. Furthermore, the right hand side of this matrix is just the left hand side polynomials evaluated at the point $1$. That is, the original determinant is equal to
$$ \begin{array}{|cccccccccc|} 1 & 0 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & 0 \\ x & M(x) & M^2(x) & \cdots & M^{m-1}(x) & M^{0}(x)(1) & M(x)(1) & M^2(x)(1) & \cdots & M^{m-1}(x)(1) \\ x^{2} & M(x^{2}) & M^{2}(x^{2}) & \cdots & M^{m-1}(x^{2}) & M^0(x^2)(1) & M(x^2)(1) & M^2(x^2)(1) & \cdots & M^{m-1}(x^2)(1) \\ x^{3} & M(x^{3}) & M^{2} (x^{3}) & \cdots & M^{m-1}(x^{3}) & M^0(x^3)(1) & M(x^3)(1) & M^{2}(x^3)(1) & \cdots & M^{m-1}(x^3)(1) \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ x^{n} & M(x^{n}) & M^{2}(x^{n}) & \cdots & M^{m-1}( x^{n}) & M^0(x^n)(1) & M(x^n)(1) & M^2(x^n)(1) & \cdots & M^{m-1}(x^n)(1) \end{array} $$
Before executing the necessary transformation, we will set $M^0(p(x)) = p(x)$ (the identity transformation) and make the following simple observations to be used in this section :
$M^k$ is a linear map i.e. $M^k(p(x)+q(x)) = M^k(p(x)) + M^k(q(x))$ for any $p,q$ polynomials and $k \geq 1$. Further, $M^k(cp(x)) = cM^k(p(x))$ for any $k \geq 1$, polynomial $p$ and real constant $c$.
$M(p(x)) = M(p(x)+c)$ for any constant $c$. That's because we're differentiating before multiplying by $x$ as per the definition of $M$, and differentiation kills constants.
We will make more observations in the next section, but great so far! Now, let's do the necessary transformation. Call the rows of this matrix as $R_1,R_2,\ldots,R_n$. We will do the following transformations in this order.
$R_n \mapsto T_n(R_n, R_{n-1},R_{n-2},\ldots,R_1)$.
$R_{n-1} \mapsto T_{n-1}(R_{n-1}, R_{n-2},\ldots,R_1)$.
$R_{n-2} \mapsto T_{n-2}(R_{n-2},R_{n-3},\ldots,1)$.
$\vdots$
$R_{2} \mapsto T_2(R_2,R_1)$.
But why are we doing this? Just focus on the first entry of each row while we do this. After doing $R_n \mapsto T_n(R_n, R_{n-1},R_{n-2},\ldots,R_1)$, the first entry of $R_n$ is $$ T_n(x^n,x^{n-1},\ldots,1) = (x-1)^n. $$ After this, once you do $R_{n-1} \mapsto T_{n-1}(R_{n-1}, R_{n-2},R_{n-3},\ldots,R_1)$, the first entry of $R_{n-1}$ is $$ T_{n-1}(x^{n-1},x^{n-2},\ldots,1) = (x-1)^{n-1} $$ and so on. Thus, what we've done is a transformation that attempts to get those powers of $(x-1)$ that we sought as entries of the first row.
Let us give some special names to some polynomials generated in this process. By the definition of $T_l$, the polynomials $T_l(x^l,x^{l-1},\ldots,1)$ and $T_l(x^{l},x^{l-1},\ldots,0)$ differ only by a constant (either $1$ or $-1$ depending upon the parity of $l$). In particular, it follows that $$ M(T_l(x^{l},x^{l-1},\ldots,0)) = M(T_l(x^{l},x^{l-1},\ldots,1)) = M((x-1)^l) $$ for all $l \geq 1$ by the second stated property of $M$. Let $$ Q_l = M(T_l(x^l,x^{l-1},\ldots,0)) = M(T_l(x^{l},\ldots,1)) = M((x-1)^l)) $$
Once we complete the transformations, by linearity of $M$ and tracking stuff carefully we see that the resulting matrix is
$$ \begin{array}{|cccccccccc|} 1 & 0 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & 0 \\ (x-1) & Q_1 & M(Q_1) & \cdots & M^{m-2}(Q_1) & 0 & Q_1(1) & M(Q_1)(1) & \cdots & M^{m-2}(Q_1)(1) \\ (x-1)^2 & Q_2 & M(Q_2) & \cdots & M^{m-2}(Q_2) & 0 & Q_2(1) & M(Q_2)(1) & \cdots & M^{m-2}(Q_2)(1) \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ (x-1)^n & Q_n & M(Q_n) & \cdots & M^{m-2}(Q_n) & 0 & Q_n(1) & M(Q_n)(1) & \cdots & M^{m-2}(Q_n)(1) \end{array} $$
It is time for us to now understand the matrix here, but we want to particularly understand the right hand side of this matrix. In order to do this, we supplement ourselves with the following fact about $M$
This follows by the product rule for differentiation and induction (or the Leibniz rule for higher derivatives).
Great. Now, remember that if a polynomial has $x-1$ as a factor, then the polynomial evaluated at $1$ is $0$. Therefore, by the first property of $M$ above, it follows that $M^k(Q_l)(1) = M^{k+1}((x-1)^l)(1) = 0$ whenever $l>k+1$.
However, this makes quite a massive difference to the matrix. Indeed, suppose that we divide the above matrix into its upper left, upper right, lower left and lower right $m \times m$ submatrices. Call the resulting subdivision $\begin{bmatrix}A&B\\C&D\end{bmatrix}$. As a result of our observations,
$D \equiv 0$ : look at the description of $D$ and our result ($M^{k+1}((x-1)^l)(1) = 0$ whenever $l>k+1$) for this.
$B$ is an upper triangular matrix. Indeed, its lower triangle completely vanishes because of our result again. ($B$ also has non-zero constant diagonal entries, but we don't need to worry about that).
Now, imagine doing the Laplace expansion of a matrix of the form $\begin{bmatrix}A&B\\C&\mathbf{0}\end{bmatrix}$ where $B$ is upper triangular. Let's take an example by actually showing what happens in the case $m=2$ for us, after doing the reductions : $$ \begin{bmatrix} 1&0 & 1 & 0 \\ (x-1)&x & 0 & 1 \\ (x-1)^2& 2x(x-1)&0&0\\ (x-1)^3&3x(x-1)^2&0&0\\ \end{bmatrix} $$ Can you see what happens? Indeed, when we do the Laplace expansion, because $B$ is upper triangular, the only way in which we obtain non-zero terms of the expansion is when we include all the diagonal entries of $B$, for otherwise, one of the lower-triangular entries of $B$ would enter the expansion, but then that's just $0$!
However, all such terms of the Laplace expansion completely ignore $A$, and the rest of such a term would consist of some entries of $C$. Basically, what ends up happening is that :
$$ \det \begin{bmatrix} A & B \\ C & \mathbf{0} \end{bmatrix} = \det(B)\det(C) $$ when $B$ is upper triangular! This can easily be proven using the Laplace expansion itself. Furthermore, $\det B$ is equal to the product of the diagonal entries of $B$.
Thus, recognizing what $B$ and $C$ are in our case, the original determinant is equal to $$ (1 \times Q_2(1) \times M(Q_3)(1) \times M^2(Q_4)(1) \times\ldots \times M^{m-2}(Q_m)(1)) \\ \times \det \begin{vmatrix} (x-1)^m & Q_{m} & M(Q_m) & \ldots & M^{m-2}(Q_m) \\ (x-1)^{m+1} & Q_{m+1} & M(Q_{m+1}) & \ldots & M^{m-2}(Q_{m+1}) \\ \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ (x-1)^n & Q_n & M(Q_n) & \ldots & M^{m-2}(Q_n) \end{vmatrix} $$
The first is just a bunch of real numbers, so we have to show that the determinant of the given matrix is divisible by our polynomial. However, observe that
All entries in the first column are divisible by $(x-1)^m$.
All entries in the 2nd column are divisible by $x$, because $Q_l = M((x-1)^l)$. Furthermore, they are all also divisible by $(x-1)^{m-1}$ because of our rule on the divisibility of $M(p(x))$ by $(x-1)^j$.
All entries in the 3rd column are divisible by $x^2$ and $(x-1)^{m-2}$,
and so on. Basically, all entries in the $l$th column are divisible by $x^{l-1}(x-1)^{m-1+l}$, for $1 \leq l \leq m$. As a consequence of this fact and usual determinant rules, the entire determinant is divisible by $$x^{\sum_{l=1}^{m} (l-1)} (x-1)^{\sum_{l=1}^m (m-l+1)}= x^{(m^2-m)/2}(x-1)^{(m^2+m)/2}.$$
Let's just divide out by those entries, because there's still more fun to be had. $$ \det \begin{vmatrix} (x-1)^m & Q_{m} & M(Q_m) & \ldots & M^{m-2}(Q_m) \\ (x-1)^{m+1} & Q_{m+1} & M(Q_{m+1}) & \ldots & M^{m-2}(Q_{m+1}) \\ \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ (x-1)^n & Q_n & M(Q_n) & \ldots & M^{m-2}(Q_n) \end{vmatrix} \\ = x^{(m^2-m)/2}(x-1)^{(m^2+m)/2} \det \begin{vmatrix} 1 & \frac{Q_{m}}{(x-1)^{m-1}x} & \frac{M(Q_m)}{(x-1)^{m-2}x^2} & \ldots & \frac{M^{m-2}(Q_m)}{(x-1)x^{m-1}} \\ (x-1) & \frac{Q_{m+1}}{(x-1)^{m-1}x} & \frac{M(Q_{m+1})}{(x-1)^{m-2}x^2} & \ldots & \frac{M^{m-2}(Q_{m+1})}{(x-1)x^{m-1}} \\ \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ (x-1)^{m-1} & \frac{Q_n}{(x-1)^{m-1}x} & \frac{M(Q_n)}{(x-1)^{m-2}x^2} & \ldots & \frac{M^{m-2}(Q_n)}{(x-1)x^{m-1}} \end{vmatrix} $$
I leave the reader to see using previous arguments that every entry of the $l$th row is divisible by $(x-1)^{l-1}$, $l=1,2,\ldots,m$. All in all, we obtain yet another factor of $(x-1)^{\sum_{l=1}^{m} (l-1)} = (x-1)^{(m^2-m)/2}$.
Combining our work, we have proved that the determinant in question is divisible by $x^{(m^2-m)/2}(x-1)^{m^2}$. However, this means that the determinant has degree at least $m^2+\frac{m^2-m}{2}$, which equals the upper bound we derived.
All in all the determinant must be of the form $Cx^{(m^2-m)/2}(x-1)^{m^2}$, where $C$ is some unknown constant to be found.
Step 3
How do we find this constant? The idea is to note that if the determinant is $Cx^{(m^2-m)/2}(x-1)^{m^2}$ then every coefficient of a monomial in the determinant is $C$ times the coefficient of the same monomial in $x^{(m^2-m)/2}(x-1)^{m^2}$. Therefore, the choice of a convenient monomial will help us. But which monomial is convenient?
One suitable answer is the highest degree monomial $x^{m^2+(m^2-m)/2}$, which we considered at the start of this answer (Those were the days!). Let's see how this works. Recall the definition of the determinant via the Laplace expansion : for an $n \times n$ matrix $A = (a_{i,j})_{1 \leq i,j \leq n}$, let $S_n$ denote the set of all permutations of $\{1,\ldots,n\}$ and $\mbox{sgn}(\sigma)$ denote the sign of $\sigma$ as a permutation. Then, $$ \det A = \sum_{\sigma \in S_n} \mbox{sgn}(\sigma)\prod_{i=1}^{n} a_{i,\sigma(i)}. $$ Now, let's consider our original matrix $$ \begin{array}{|cccccccccc|} 1 & 0 & 0 & \cdots & 0 & 1 & 0 & 0 & \cdots & 0 \\ x & x & x & \cdots & x & 1 & 1 & 1 & \cdots & 1 \\ x^{2} & 2 x^{2} & 2^{2} x^{2} & \cdots & 2^{m-1} x^{2} & 1 & 2 & 2^{2} & \cdots & 2^{m-1} \\ x^{3} & 3 x^{3} & 3^{2} x^{3} & \cdots & 3^{m-1} x^{3} & 1 & 3 & 3^{2} & \cdots & 3^{m-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ x^{n} & n x^{n} & n^{2} x^{n} & \cdots & n^{m-1} x^{n} & 1 & n & n^{2} & \cdots & n^{m-1} \end{array} $$ The question is : which $\sigma$ lead to a term that contains the highest possible power $x^{m^2+(m^2-m)/2}$? The answer, by simple observation, is : any permutation that ensures that the set $\{(m+1),(m+2),\ldots,n+1\}$ is mapped to the set $\{1,2,\ldots,m\}$, so that a row numbered greater than $m$ is mapped to a column number between $1$ and $m$. This then captures all those powers in the bottom left corner as well as possible.
Let $S = \{\sigma : \sigma \text{ maps }\{(m+1),(m+2),\ldots,n+1\} \text{ to } \{1,2,\ldots,m\}\}$. Then, the coefficient of the monomial $x^{m^2+(m^2-m)/2}$ in the determinant is precisely equal to $$ x^{- (m^2+(m^2-m)/2)} \times \sum_{\sigma \in S} \mbox{sgn}(\sigma)\prod_{i=1}^{n} A_{i,\sigma(i)}, $$ where $A$ is the matrix considered at the start of this problem. But here, notice that $\sigma \in S$ if and only if essentially breaks up into two separate permutations : one from $\{(m+1),(m+2),\ldots,n+1\}$ to $\{1,2,\ldots,m\}$ and the other from $\{1,2,\ldots,m\}$ to $\{(m+1),(m+2),\ldots,n+1\}$. That is, $\sigma \in S$ if and only if there exist two permutations $\sigma_1,\sigma_2$ of $\{1,2,\ldots,m\}$ such that $$ \sigma(k) = \begin{cases} \sigma_1(k+m) & 1 \leq k \leq m \\ \sigma_2(k-m) & m-1 \leq k \leq 2m(= n+1) \\ \end{cases} $$ Furthermore, it's easy to see that $\mbox{sgn}(\sigma) = \mbox{sgn}(\sigma_1)\mbox{sgn}(\sigma_2)$ by writing $\sigma$ as a composition of extended versions of $\sigma_1$ and $\sigma_2$ to $\{1,2,\ldots,2m\}$. Therefore, the coefficient in question is equal to \begin{align*} &x^{- (m^2+(m^2-m)/2)} \times \sum_{\sigma_1,\sigma_2 \in S_m} \mbox{sgn}(\sigma_1)\mbox{sgn}(\sigma_2) \prod_{i=1}^{m} A_{i,\sigma_1(i)}\prod_{i=m+1}^{n} A_{i,\sigma_2(m-i)}\\ =& x^{- (m^2+(m^2-m)/2)} \times \sum_{\sigma_1 \in S_m} \mbox{sgn}(\sigma_1)\prod_{i=1}^{m} A^{tr}_{i,\sigma_1(i)} \times \sum_{\sigma_2 \in S_m} \mbox{sgn}(\sigma_2)\prod_{i=1}^{m} A^{bl}_{i,\sigma_2(i)}, \end{align*} where $A^{tr}$ is the top right $m \times m$ submatrix of $A$ and $A^{bl}$ is the bottom-right $m \times m$ submatrix of $A$. However, in the second of these sums, every term is divisible by $x^{- (m^2+(m^2-m)/2)}$, so at the end of the day, recognizing the Laplace expansion in each sum above, we just get that the coefficient is equal to $$ \det(A^{tr}) \times x^{- (m^2+(m^2-m)/2)} \det(A^{bl}) \\ = \det \begin{bmatrix} 1&0&0&\ldots&0\\ 1&1&1&\ldots&1\\ 1&2^1&2^2&\ldots&2^{m-1}\\ 1&3^1&3^2&\ldots&3^{m-1}\\ & \vdots & \vdots & \vdots & \vdots \\ 1&(m-1) & (m-1)^2&\ldots&(m-1)^{m-1} \end{bmatrix} \times \det \begin{bmatrix} 1&m&m^2&\ldots&m^{m-1}\\ 1&(m+1)&(m+1)^2&\ldots&(m+1)^{m-1}\\ 1&(m+2)&(m+2)^2&\ldots&(m+2)^{m-1}\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ 1&n&n^2&\ldots&n^{m-1} \end{bmatrix} $$
We know these matrices though : they're Vandermonde matrices! Their determinant is super well-known, and is equal to the product of the pairwise differences of every number whose power is being taken. So for example, in the first matrix we're taking the powers of $0,1,2$ up to $m-1$ so the answer is $$ (1-0)(2-0)(2-1)(3-0)(3-1)(3-2)\ldots ((m-1)-(m-2)) = 1!2!\ldots (m-1)! = \prod_{i=1}^{m-1} i! $$ However, the second matrix also has a similar difference pattern and yields the same product $\prod_{i=1}^{m-1} i!$. We conclude that the coefficient of $x^{m^2+(m^2-m)/2}$ equals $\left(\prod_{i=1}^{m-1} i!\right)^2$.
Finally, as the soap opera draws to a close after three generations of strife, we conclude that $$ \boxed{\det A = (x-1)^{m^2}x^{(m^2-m)/2} \left(\prod_{i=1}^{m-1} i!\right)^2} $$
as desired.