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!

Another way to the formula: I think that you have almost finished the job.
I consider your formula with the $v_k$. Replace them by $w_1,..,w_n$ any vectors in $E=\mathbb{R}^n$. Your formula give then an application $f(w_1,..w_n)$ from $E^n$ to $\mathbb{R}$, and you have see that this is $a{\rm det}(w_1,...,w_n)$ for a constant $a$, depending on the $x_k$, but not on the $w_k$. Now to find $a$, you can take for the $w_k$, the vectors of the canonical basis of $E$, and it is easy to finish.
