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  "body": "\n## Proof\n\nWe compare coefficients in a polynomial identity in an extra variable $t$. This is the generating-function proof, cut off at a finite degree so that only polynomials appear.\n\n1. Let $R$ be a [[commutative ring|0GYGHKX834]], let $n \\ge 1$, and let $A = R[x_1, \\dots, x_n]$ be the [[polynomial ring|0CES91M79A]]. We work in the [[polynomial ring|0Q9EZS90GY]] $A[t]$.\n2. Let $e_k$ be the [[elementary symmetric polynomials|0ZMY8441GW]], with $e_0 = 1$ and $e_k = 0$ for $k \u003e n$. Let $p_k$ be the [[power sums|0AS1FZT6BM]].\n3. Let $E(t) = \\prod_{i=1}^n (1 + x_i t)$. Expanding, choose $1$ or $x_i t$ from each factor. A choice is a subset $S \\subseteq \\set{1, \\dots, n}$, and it contributes $t^{\\abs{S}} \\prod_{i \\in S} x_i$. So $E(t) = \\sum_{j \\ge 0} e_j t^j$.\n4. For each $i$, let $E_i(t) = \\prod_{j \\neq i} (1 + x_j t)$. In the same way, $E_i(t) = \\sum_{j \\ge 0} e^{(i)}_j t^j$. Here $e^{(i)}_j$ is the sum of the products of $j$ distinct variables other than $x_i$.\n5. Fix $k \\ge 1$. For each $i$, let $Q_i(t) = \\sum_{m=1}^{k} (-1)^{m-1} x_i^m t^m$.\n6. The product $(1 + x_i t) Q_i(t)$ telescopes: the term $(-1)^{m-1} x_i^m t^m$ of $Q_i$ cancels against $x_i t$ times the term with $m - 1$. So $(1 + x_i t) Q_i(t) = x_i t + (-1)^{k-1} x_i^{k+1} t^{k+1}$.\n7. Since $E(t) = (1 + x_i t) E_i(t)$, step 6 gives $E(t) Q_i(t) = x_i t E_i(t) + (-1)^{k-1} x_i^{k+1} t^{k+1} E_i(t)$.\n8. Summing over $i$, and using $\\sum_i Q_i(t) = \\sum_{m=1}^{k} (-1)^{m-1} p_m t^m$, we get\n$$E(t) \\sum_{m=1}^{k} (-1)^{m-1} p_m t^m = \\sum_{i=1}^n x_i t E_i(t) + t^{k+1} H(t)$$\nfor some $H(t) \\in A[t]$.\n9. The coefficient of $t^k$ on the left is $\\sum_{m=1}^{k} (-1)^{m-1} e_{k-m} p_m$, by step 3. The term $t^{k+1} H(t)$ has no $t^k$ term.\n10. By step 4, the coefficient of $t^k$ in $\\sum_i x_i t E_i(t)$ is $\\sum_{i=1}^n x_i e^{(i)}_{k-1}$.\n11. Each term of $x_i e^{(i)}_{k-1}$ is $\\prod_{j \\in S} x_j$ for a $k$-element set $S$ with $i \\in S$. So each $k$-element set $S$ appears once for each $i \\in S$, that is, $k$ times. Hence $\\sum_i x_i e^{(i)}_{k-1} = k \\, e_k$.\n12. This also holds for $k \u003e n$, where both sides are $0$, since there are no $k$-element subsets.\n13. Comparing the coefficients of $t^k$ in step 8 gives $k \\, e_k = \\sum_{i=1}^{k} (-1)^{i-1} e_{k-i} \\, p_i$ for every $k \\ge 1$.\n\n## Idea\n\nThe polynomial $E(t) = \\prod (1 + x_i t)$ has the $e_k$ as coefficients. Multiplying it by an alternating sum of power sums gives $t$ times the derivative of $E(t)$, up to terms of high degree. The coefficient of $t^k$ in $t E'(t)$ is $k e_k$, which gives the identity.\n",
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