In my earlier (soft) MO post, an elementary response was given by Ofir Gorodetsky in regard to the determinant of the symbolic counterpart to the numerical matrix $\mathbf{M}_n=(i^j-j^i)_{i,j}^{1,n}$.
I'm now looking at yet another variation. Let $\mathbf{A}_n=(j^j-i^j)_{i,j}^{1,n}$ be an $n\times n$-matrix. Denote the (signed) Stirling numbers of the first kind by $s(n,k)$. For clarity, here are some contrasting examples:
$$\mathbf{M}_3=\begin{pmatrix} 0&-1&-2 \\ 1&0&-1 \\2&1&0 \end{pmatrix}
\qquad \text{and} \qquad \mathbf{A}_3=\begin{pmatrix}0&3&26 \\-1&0&19 \\ -2&-5&0 \end{pmatrix}.$$
I would like to ask:
QUESTION. Is this true? With the convention that $0^0=1$, we have
$$\det\mathbf{A}_n=\prod_{j=1}^{n-1}j!\cdot\sum_{k=0}^n\,s(n+1,k+1)\cdot k^k.$$
Remark. If it helps, since the factor $\prod_{j=1}^{n-1}j!$ is exactly the determinant of the Vandermonde matrix, $\mathbf{V}_n=(i^{j-1})_{i,j}^{1,n}$, we can say
$$\frac{\det\mathbf{A}_n}{\det \mathbf{V}_n}=\sum_{k=0}^n\,s(n+1,k+1)\cdot k^k.$$ This formalism goes in the "spirit" of the (specialization) $s_{\lambda}(1,\cdots,1)$ of the Schur polynomials.
Best Answer
This is true and this is not about numbers $j^j$ at all. Consider a more genral matrix $(c_j-i^j)_{1\leqslant i,j\leqslant n}$. Denote $f_j(x)=c_j-x^j$, and find the numbers $\beta$, $\alpha_1,\ldots,\alpha_{n-1}$ such that $f_n(x)+\sum_{j=1}^{n-1}\alpha_j f_j(x)=\beta$ for all $x\in \{1,2,\ldots,n\}$. Then the matrix $(f_j(i))_{i,j}$ has the same determinant as if we replace the column with $f_n$ by a column of $\beta$'s, that is seen from operations with columns. Now by operations with columns we may replace each $f_j$ to $-x^j$, so we get almost a Vandermonde matrix (up to $n-1$ changes of signs in the columns and the cyclic shift, and $\beta$'s instead of 1's), so its determinant equals to the determinant of Vandermonde matrix for the numbers $1,\ldots,n$ times $(-1)^{n-1}(-1)^{n-1}\beta=\beta$. Thus, it appears to find $\beta$. For this, write $$ f_n+\sum_{j=1}^{n-1}\alpha_jf_j(x)=\beta-(x-1)(x-2)\ldots(x-n)= \beta-\sum_{k=0}^n s(n+1,k+1)x^k. $$ Equalising the coefficients of $x^k$ we get $\alpha_k=s(n+1,k+1)$ for all $k=1,\ldots,n$ (where $\alpha_n=1$ by agreement), thus $\beta-s(n+1,1)=\sum_{k=1}^n c_k s(n+1,k+1)$, $\beta=\sum_{k=0}^n c_k s(n+1,k+1)$, where $c_0=1$ by agreement.