# Determinant with entries from the sum $s_k = x_1^k + \cdots + x_n^k$?

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?

$$\begin{bmatrix} 1 & \cdots & 1 & 1\\ x_1 & \cdots & x_{n} & y\\ x_1^2 & \cdots & x_{n}^2 & y^2 \\ \vdots & \cdots & \vdots &\vdots \\ x_1^n & \cdots & x_{n}^n & y^n \end{bmatrix} \cdot \begin{bmatrix} 1 & x_1 & \dots & x_1^{n-1} & 0\\ 1 & x_2 & \dots & x_2^{n-1} & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 1 & x_n & \dots & x_n^{n-1} & 0 \\ 0 & 0 & \dots & 0 & 1 \end{bmatrix} = \begin{bmatrix} s_0 & s_1 & \cdots & s_{n-1} & 1\\ s_1 & s_2 & \cdots & s_{n} & y\\ \vdots & \vdots & \vdots & \ddots &\vdots \\ s_n & s_{n+1} & \cdots & s_{2n-1} & y^n \end{bmatrix}$$
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 $$\det(AB) = \det(A)\det(B)$$.