Any neat way to calculate this Vandermonde-like determinant? Let $x_i,i\in\{1,\cdots,n\}$ be real numbers, and $s_k=x_1^k+\cdots+x_n^k$, I'm asked to calculate
$$
    |S|:=
    \begin{vmatrix}
      s_0 & s_1 & s_2 & \cdots & s_{n-1}\\
      s_1 & s_2 & s_3 & \cdots & s_n\\
      s_2 & s_3 & s_4 & \cdots & s_{n+1}\\
      \vdots & \vdots & \vdots & \ddots & \vdots \\
      s_{n-1} & s_{n} & s_{n+1} & \cdots & s_{2n-2}
    \end{vmatrix}
$$
and to prove that $|S|\ge 0$ for all possible real $x_i$.
I found that
$$
|S|=\det[(v_1+\cdots v_n), (x_1v_1+\cdots+x_nv_n),\cdots,(x_1^{n-1}v_1+\cdots+x_n^{n-1}v_n)],\quad\text{where}\, v_j=\begin{bmatrix}
  1 \\ x_j \\ \vdots\\ x_j^{n-1}
\end{bmatrix}
$$
Due to multilinearity of the $\det$ function, I sense it might have something to do with Vandermonde determinant. In fact, it must have the form
$$|S|=(\det[v_1,\cdots, v_n])\cdot \text{something}$$
But that "something" involves many cyclic sums and is therefore a horrible mess..
Anyway, is there a neat way to calculate this tricky determinant? Thanks!

Hint:
$$\begin{bmatrix}1&1&1\\x_1&x_2&x_3\\x_1^2&x_2^2&x_3^2\end{bmatrix}\begin{bmatrix}1&x_1&x_1^2\\1&x_2&x_2^2\\1&x_3&x_3^2\end{bmatrix}=\begin{bmatrix}1+1+1&x_1+x_2+x_3&x_1^2+x_2^2+x_3^2\\x_1+x_2+x_3&x_1^2+x_2^2+x_3^2&x_1^3+x_2^3+x_3^3\\x_1^2+x_2^2+x_3^2&x_1^3+x_2^3+x_3^3&x_1^4+x_2^4+x_3^4\end{bmatrix}$$
