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  "body": "\n## Proof\n\n1. Let $p$ be prime, let $n \\ge 1$ and let $q = p^n$. Let $f = x^q - x \\in \\F_p[x]$.\n2. By the [[existence of splitting fields|01441KDGBY]], $f$ has a [[splitting field|02NZZ1J0FH]] $L$ over $\\F_p$. So $f = (x - a_1) \\cdots (x - a_q)$ with $a_i \\in L$, since $f$ is monic of degree $q$ and [[splits|0SHMC00RJV]] over $L$.\n3. The field $L$ contains $\\F_p$, so $p \\cdot 1 = 0$ in $L$, and the [[characteristic|0P7CH1CND5]] of $L$ is $p$.\n4. The [[formal derivative|0AD4VS02H3]] of $f$ is $f' = q x^{q-1} - 1$. Since $q \\cdot 1 = p^{n-1} (p \\cdot 1) = 0$ in $L$, we get $f' = -1$.\n5. Suppose that some root $c$ of $f$ in $L$ has [[multiplicity|03XVP8Q4GY]] at least $2$. Then $f'(c) = 0$ by the [[criterion for repeated roots|067F8YVYMH]]. This contradicts $f'(c) = -1 \\neq 0$.\n6. So every root of $f$ has multiplicity $1$, and $a_1, \\dots, a_q$ are distinct. Let $R = \\set{a_1, \\dots, a_q}$, so $\\abs{R} = q$ and $R = \\set{a \\in L : a^q = a}$.\n7. The [[Frobenius endomorphism|0GHDDVAH8P]] $\\varphi(a) = a^p$ of $L$ is a ring homomorphism, by the [[freshman's dream|0396VGFA78]]. So $\\psi = \\varphi^n$, $\\psi(a) = a^{q}$, is a ring homomorphism $L \\to L$.\n8. So $R = \\set{a \\in L : \\psi(a) = a}$ contains $0$ and $1$. For $a, b \\in R$, we get $\\psi(a - b) = \\psi(a) - \\psi(b) = a - b$ and $\\psi(ab) = \\psi(a)\\psi(b) = ab$.\n9. For $a \\in R$ with $a \\neq 0$, we get $\\psi(a^{-1}) = \\psi(a)^{-1} = a^{-1}$, since $\\psi(a)\\psi(a^{-1}) = \\psi(1) = 1$.\n10. By steps 8 and 9, $R$ is a [[subfield|07QY5RPFBR]] of $L$. By step 6 it has $q = p^n$ elements, so it is a [[finite field|08RTDHX1SK]] of order $p^n$.\n\n## Idea\n\nBuild a field in which $x^q - x$ splits, and look at its roots. The derivative is the constant $-1$, so the roots are distinct and there are exactly $q$ of them. The map $a \\mapsto a^q$ is a ring homomorphism in characteristic $p$, so its fixed points form a field.\n",
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