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  "body": "\n## Proof\n\n1. Let $F$ be a [[finite field|08RTDHX1SK]] of order $q = p^n$, with $p$ prime. By the [[prime power theorem|0H5WWC8RZX]], $q = p'^{n'}$, where $p'$ is the characteristic of $F$. By unique factorisation of $q$, $p' = p$.\n2. Let $P$ be the [[prime subfield|0VPFBAY7MW]] of $F$. It is the copy $\\F_p$ of the multiples of $1$, as in that theorem.\n3. The polynomial $f = x^q - x$ has coefficients $1$, $-1$ and $0$, so $f \\in P[x]$, and $F/P$ is a field extension.\n4. By the [[factorisation of $x^q - x$|0QWR8WTZEW]], $f = \\prod_{a \\in F} (x - a)$ in $F[x]$. So $f$ [[splits|0SHMC00RJV]] over $F$, and its roots are exactly the elements of $F$.\n5. The [[field generated|0MA5B744K3]] over $P$ by the roots of $f$ is $P(F)$. This is the smallest subfield of $F$ that contains $F$, so it is $F$.\n6. By steps 4 and 5, $F$ satisfies both conditions of a [[splitting field|02NZZ1J0FH]] of $x^q - x$ over $\\F_p$.\n\n## Idea\n\nEvery element of $F$ is a root of $x^q - x$, and this polynomial has no other roots. So it splits over $F$, and its roots already make up the whole field.\n",
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