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equals the determinant of the matrix obtained by substituting $a_i$ for $x$, so from the original matrix $a_1$ has been replaced by $a_i$, and as this matrix has its rows $1$ and $i$ identical, its determinant vanishes.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
$$
V(a_1,a_2,\ldots,a_n)
=\left(\prod_{2\leq i<j\leq n}(a_j-a_i)\right)\prod_{j=2}^n(a_j-a_1)
=\prod_{1\leq i<j\leq n}(a_j-a_i).
$$
Hint. Rewrite left part using complex numbers:
$$
\prod_{j=1}^n\left(1+x_j^2\right)=
\prod_{j=1}^n\left[\left(1+\textbf i \:x_j\right)\left(1-\textbf i \:x_j\right)\right]
$$
Then try to use Vieta's formulas. General idea is to find the way to use $x_j$ instead of $x_j^2$.
UPD (using @Exodd hint):
Let's define the following function:
$$
g(z):=\prod_{j=1}^n\left(1-\frac{x_j}{z}\right)=
\prod_{j=1}^n\frac{z-x_j}{z}=\frac{1}{z^n}\prod_{j=1}^n\left(z-x_j\right)=
\frac{f(z)}{z^n}
$$
Then
$$
\prod_{j=1}^n\left(1+x_j^2\right)=
\prod_{j=1}^n\left[\left(1+\textbf i \:x_j\right)\left(1-\textbf i \:x_j\right)\right]=
\prod_{j=1}^n\left(1+\textbf i\:x_j\right)\cdot\prod_{j=1}^n\left(1-\textbf i\:x_j\right)=
$$
$$
=\prod_{j=1}^n\left(1-\frac{x_j}{\textbf i}\right)\cdot\prod_{j=1}^n\left(1+\frac{x_j}{\textbf i}\right)=
g(\textbf i)g(-\textbf i)
$$
Now we notice that $g(\bar z)=\overline{g(z)}$, so
$$
g(\textbf i)g(-\textbf i) = |g(\textbf i)|^2=
\left|\frac{f(\textbf i)}{\textbf i^n}\right|^2 =
\left|f(\textbf i)\right|^2
$$
That's it. Not to deal with some technical stuff (general $n$ case, a lot of indices...) let's show for small $n=4$:
$$
\left|f(\textbf i)\right|^2=
\left|\textbf i^4+a_1\textbf i^3+a_2\textbf i^2+a_3\textbf i+a_4\right|^2=
$$
$$
=\left|1-a_1\:\textbf i-a_2+a_3\:\textbf i+a_4\right|^2=
\left|(1-a_2+a_4)-\textbf i\:(a_1-a_3)\right|^2=
$$
$$
=(1-a_2+a_4)^2+(a_1-a_3)^2
$$
Best Answer
Let $$P(x) = a_nx^n+....+a_1x+a_0,$$ then we have $$ a_nx^n + ...+a_1x+a_0 = (a_n\cdot n\cdot a) x^n+...+(aa_1+2a_2b)x+a_1b$$ Since $a_n\neq 0$ we get $an=1$, so $a=1/n$
Write $$ {P'(x) \over P(x)} = {1\over ax+b}$$
Then $$a(\ln P(x))' = (\ln(ax+b))'$$ and thus $$a\ln(P(x) = \ln(ax+b) + c$$
So $$P(x) = (ax+b)^{1\over a} e^c = (ax+b)^ne^c$$