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  "body": "\n## Proof\n\n1. Let $F$ be a [[finite field|08RTDHX1SK]] with $q$ elements. Let $F^\\times$ be its [[group of units|05THZKR9AN]].\n2. In a field, every nonzero element is a unit. So $F^\\times = F \\setminus \\set{0}$, and $F^\\times$ is a group of order $q - 1$.\n3. Let $a \\in F$ with $a \\neq 0$. Then [[$a^{q-1} = 1$|0ZEPNS9BXK]], since $a$ lies in the finite group $F^\\times$ of order $q - 1$.\n4. Multiplying by $a$ gives $a^q = a$.\n5. For $a = 0$ we have $a^q = 0 = a$, since $q \\ge 1$. So $a^q = a$ for every $a \\in F$.\n6. Let $f = x^q - x$ and $g = \\prod_{a \\in F} (x - a)$ in $F[x]$. Both are monic of [[degree|0DPF6WRZW6]] $q$.\n7. So the leading terms cancel in $h = f - g$, and $h$ is $0$ or has degree less than $q$.\n8. Every $a \\in F$ is a root of $f$ by step 5. It is also a root of $g$, because $g$ has the factor $x - a$. So every $a \\in F$ is a root of $h$.\n9. Thus $h$ has $q$ distinct roots in $F$. A nonzero polynomial of degree less than $q$ has [[fewer than $q$ roots|0KJVQV3E0F]]. So $h = 0$.\n10. Hence $x^q - x = \\prod_{a \\in F} (x - a)$ in $F[x]$.\n\n## Idea\n\nThe nonzero elements form a group of order $q - 1$, so Lagrange's theorem gives $a^{q-1} = 1$ for them. Then all $q$ elements are roots of $x^q - x$. A monic polynomial of degree $q$ with $q$ known distinct roots is the product of the linear factors, because the difference would have too many roots.\n",
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