For part (a), this is just development (Laplace expansion) of the determinant by the first row. Actually the $\det()$ factors should have alternating signs. Since the only occurrences of $x$ are in that first row, all the $\det()$ expressions are constants, and one gets a polynomial of degree at most $n-1$ (from the final term) in $x$.
For part (b), that $P(a_i)=0$ for $i=2,3,\ldots,n$ is just the fact that $P(a_i)$ equals the determinant of the matrix obtained by substituting $a_i$ for $x$, so from the original matrix $a_1$ has been replaced by $a_i$, and as this matrix has its rows $1$ and $i$ identical, its determinant vanishes. all this uses is that the Laplace expansion used commutes with such substitution. Furthermore a polynomial of degree at most $n-1$ with $n-1$ specified roots $a_2,\ldots,a_n$ can only be a scalar multiple of $(x-a_2)\ldots(x-a_n)$.
For part (c), this is just remarking that the $\det()$ in question is $(-1)^{n-1}$ times the determinant of the lower-left $(n-1)\times(n-1)$ submatrix, which determinant precisely matches the definition of $V_{n-1}(a_2,\ldots,a_n)$.
For part (d) write $(-1)^{n-1}\prod_{i=2}^n(x-a_i)=\prod_{i=2}^n(a_i-x)$ to get
$$
V(x,a_2,\ldots,a_n)=
(-1)^{n-1}V_{n-1}(a_2,\ldots,a_n)\prod_{i=2}^n(x-a_i)
=V_{n-1}(a_2,\ldots,a_n)\prod_{i=2}^n(a_i-x),
$$
and then set $x=a_1$ to get
$$
V(a_1,a_2,\ldots,a_n)
=V_{n-1}(a_2,\ldots,a_n)\prod_{i=2}^n(a_i-a_1),
$$
Part (e) applies induction on $n$ to $V_{n-1}(a_2,\ldots,a_n)$ (the starting case is $V_0()=1=\prod_{1\leq i<j\leq 0}1$, an empty product, or if you fear $n=0$ it is $V_1(a)=1=\prod_{1\leq i<j\leq 1}1$, still an empty product), to get
$$
V(a_1,a_2,\ldots,a_n)
=\left(\prod_{2\leq i<j\leq n}(a_j-a_i)\right)\prod_{j=2}^n(a_j-a_1)
=\prod_{1\leq i<j\leq n}(a_j-a_i).
$$
If you are willing to reduce $A$ to column echelon form, and accept that $\det$ is multilinear and alternating, then there is a proof that is straightforward and avoids the Leibnitz formula.
The idea is simple, just reduce $A$ to upper triangular form.
Suppose the columns of $A$ are $a_k$. We want to compute $f(a_1,...,a_n)$. Note that we have $f(a_1,...,a_k+ \lambda a_p,...,a_n) = f(a_1,...,a_n)$ for any $p \neq k$ and any $\lambda$. Since $\det$ is also multilinear and alternating, we have $\det(a_1,...,a_k+ \lambda a_p,...,a_n) = \det(a_1,...,a_n)$. Similarly, if any two columns are switched, both $f$ and $\det$ change sign.
Now switch columns and add multiples of another column to a column so that the resulting $a_1',...,a_n'$ are in column echelon form. Then we have
$f(a_1',...,a_n') = (-1)^k f(a_1,...,a_n)$ and $\det(a_1',...,a_n') = (-1)^k \det(a_1,...,a_n)$, where $k$ is the number of times a pair of columns was switched.
Since $a_1',...,a_n'$ are in column echelon form, we have
$f(a_1',...,a_n') = (-1)^k (a_1')_1 ... (a_n')_n f(e_1,...,e_n)$ and similarly, $\det(a_1',...,a_n') = (-1)^k (a_1')_1 ... (a_n')_n $. In particular, we have
$f(a_1,...,a_n) = \det(a_1,...,a_n) f(e_1,...,e_n)$.
Best Answer
I assume the matrices $A_i$ are in $\Bbb C^{d\times d}$. Let $M:=a_1A_1+\cdots+a_nA_n$ and $N:=|a_1|A_1+\cdots+|a_n|A_n$.