Prove that matrices of this form have eigenvalues $0,1,\ldots , n-1$

For $ k = 0, 1, \ldots, n-1$, consider the (horizontal) vector $v_k$ with $i$th coordinate $$ \sum \prod_{j=1, a_j \neq i}^{k} {x_{a_j}}.$$

For example, with $n = 3$, we have
$v_0 = (1, 1, 1)$,
$v_1 = (x_2 + x_3, x_3 + x_1, x_1 + x_2)$,
$v_2 = ( x_2x_3, x_3x_1, x_1x_2)$.

With $n = 4$, we have
$v_0 = (1, 1, 1, 1)$,
$v_1 = (x_2 + x_3 + x_4, x_3 + x_4 + x_1, x_4 + x_1 + x_2, x_1 + x_2 + x_3)$,
$v_2 = ( x_2x_3+x_3x_4+x_4x_2, x_3x_4+x_4x_1+x_1x_3, x_4x_1+x_1x_2+x_2x_4, x_1x_2 + x_2x_3 + x_3x_2)$,
$v_3 = (x_2x_3x_4, x_3x_4x_1, x_4x_1x_2, x_1x_2x_3)$.

Claim: $v_kA = (n-1-k) v_k$.

Proof: Expand it. A lot of the cross terms cancel out.
For example, with $v_0$, the column sum is $n-1$, so $v_0 A = (n-1) v_0$.
For example, with $v_{k-1}$, the numerators are all $\prod x_i$, and by looking at the denomninators, they cancel out to 0, so $v_{k-1} A = 0 $.

Do you see how we get $v_k A = (n-1-k)v_k$?

Corollary: The eigenvalues are $0, 1, 2, \ldots, n-1 $.