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  "body": "\n## Proof\n\nWe let a cyclic group of order $p$ act on tuples by cyclic shifts, and count fixed points modulo $p$. Let $e$ be the identity of $G$.\n\n1. Let $X$ be the set of tuples $t = (g_1, \\dots, g_p) \\in G^p$ with $g_1 g_2 \\cdots g_p = e$.\n2. The map $X \\to G^{p-1}$, $(g_1, \\dots, g_p) \\mapsto (g_1, \\dots, g_{p-1})$, is a bijection. Its inverse sets $g_p = (g_1 \\cdots g_{p-1})^{-1}$, the only element that makes the product $e$. So $\\abs{X} = \\abs{G}^{p-1}$.\n3. Since $p \\mid \\abs{G}$ and $p - 1 \\geq 1$, step 2 gives $p \\mid \\abs{X}$.\n4. Let $s(g_1, \\dots, g_p) = (g_2, \\dots, g_p, g_1)$. If $t \\in X$, then $g_2 \\cdots g_p g_1 = g_1^{-1} (g_1 g_2 \\cdots g_p) g_1 = g_1^{-1} e g_1 = e$. So $s$ maps $X$ to $X$.\n5. Shifting $p$ times returns each tuple to itself, so $s^p$ is the identity of $X$. Hence $s^k$ depends only on $k$ modulo $p$.\n6. Let $C = \\Z/p\\Z$, the [[cyclic group of order $p$|0G0B3DRG65]]. Define $k \\cdot t = s^k(t)$ for $k \\in C$ and $t \\in X$. This is well defined by step 5.\n7. We have $0 \\cdot t = t$ and $j \\cdot (k \\cdot t) = s^{j+k}(t) = (j + k) \\cdot t$. So this is an [[action|0YSE802YCQ]] of $C$ on $X$.\n8. By the [[orbit size formula|0S19CWFZM2]], each orbit has a size that divides $\\abs{C} = p$. As $p$ is prime, each orbit has size $1$ or $p$.\n9. An orbit has size $1$ exactly when its point is a [[fixed point|08JF1K4DW1]]. Let $X^C$ be the set of fixed points. Since the [[orbits partition|0ZC7FYE0NZ]] $X$, $\\abs{X} = \\abs{X^C} + p \\cdot r$, where $r$ is the number of orbits of size $p$.\n10. By steps 3 and 9, $p$ divides $\\abs{X^C}$.\n11. A tuple $t$ is fixed by all of $C$ if and only if $s(t) = t$, since every $s^k$ is a power of $s$. And $s(t) = t$ means $g_{i+1} = g_i$ for all $i \u003c p$, that is, $t = (g, g, \\dots, g)$ for some $g$.\n12. Such a constant tuple lies in $X$ if and only if $g^p = e$. So $\\abs{X^C}$ is the number of $g \\in G$ with $g^p = e$.\n13. The element $e$ satisfies $e^p = e$, so $\\abs{X^C} \\geq 1$. By step 10, $\\abs{X^C} \\geq p \\geq 2$. So there is $g \\neq e$ with $g^p = e$.\n14. Let $m$ be the [[order|0JPYSD5E0B]] of this $g$. Since [[$g^p = e$, $m$ divides $p$|0F5QC101Q7]], so $m$ is $1$ or $p$.\n15. If $m = 1$, then $g = g^1 = e$, which is false. So $m = p$, and $g$ is an element of order $p$.\n\n## Idea\n\nTuples whose product is the identity can be rotated without leaving the set, and their number $\\abs{G}^{p-1}$ is a multiple of $p$. Rotations sort the tuples into groups of $p$, except the constant tuples, which stay put. So the number of constant tuples $(g, \\dots, g)$ with $g^p = e$ is a multiple of $p$, and besides $(e, \\dots, e)$ there must be others.\n",
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