Let $s_k = x_1^k + \cdots + x_n^k$. Compute
$$\begin{vmatrix}
s_0 & s_1 & \cdots & s_{n-1} & 1\\
s_1 & s_2 & \cdots & s_{n} & y\\
\vdots & \vdots & \vdots & \ddots &\vdots \\
s_n & s_{n+1} & \cdots & s_{2n-1} & y^n
\end{vmatrix}$$
The proof "observes" that the determinant can be written as the product of the two determinants:
$$\begin{vmatrix}
1 & \cdots & 1 & 1\\
x_1 & \cdots & x_{n} & y\\ x_1^2 & \cdots & x_{n}^2 & y^2 \\
\vdots & \cdots & \vdots &\vdots \\
x_1^n & \cdots & x_{n}^n & y^n
\end{vmatrix} \cdot \begin{vmatrix}
1 & x_1 & \dots & x_1^{n-1} & 0\\
1 & x_2 & \dots & x_2^{n-1} & 0 \\
\vdots & \vdots & \cdots & \vdots &\vdots \\ 1 & x_n & \dots & x_n^{n-1} & 0 \\
0 & 0 & \dots & 0 & 1
\end{vmatrix}$$
The answer then being $\prod (y-x_i) \prod_{i > j}(x_i -x_j)^2$
My question is: where does the observation come from?
Best Answer
Notice that we can write:
$$\begin{bmatrix} 1 & \cdots & 1 & 1\\ x_1 & \cdots & x_{n} & y\\ x_1^2 & \cdots & x_{n}^2 & y^2 \\ \vdots & \cdots & \vdots &\vdots \\ x_1^n & \cdots & x_{n}^n & y^n \end{bmatrix} \cdot \begin{bmatrix} 1 & x_1 & \dots & x_1^{n-1} & 0\\ 1 & x_2 & \dots & x_2^{n-1} & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 1 & x_n & \dots & x_n^{n-1} & 0 \\ 0 & 0 & \dots & 0 & 1 \end{bmatrix} = \begin{bmatrix} s_0 & s_1 & \cdots & s_{n-1} & 1\\ s_1 & s_2 & \cdots & s_{n} & y\\ \vdots & \vdots & \vdots & \ddots &\vdots \\ s_n & s_{n+1} & \cdots & s_{2n-1} & y^n \end{bmatrix}$$
To see why this is true, notice that the dot product of the $i$-th row of the left matrix and the $j$-th column of the right matrix is $x_1^{i-1} \cdot x_1^{j-1}+\cdots+x_n^{i-1}\cdot x_n^{j-1}+y^{i-1}\cdot 0 = s_{i+j-2}$ for $j \neq n$, and $x_1^{i-1}\cdot 0 + \cdots x_n^{i-1} \cdot 0 + y^{i-1} \cdot 1 = y^{i-1}$ for $j = n$.
The result then follows from using the identity $\det(AB) = \det(A)\det(B)$.