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  "body": "\n## Proof\n\n1. Let $p$ be a [[prime number|0SN0MAEKYX]] and let $a$ be an integer with $p \\nmid a$.\n2. The only positive divisors of $p$ are $1$ and $p$, and $p \\nmid a$. So $\\gcd(a, p) = 1$, and $a$ is [[coprime|0GFKP8F8AG]] to $p$.\n3. So the [[multiplicative order|0SBRX5CZ5R]] $k = \\operatorname{ord}_p(a)$ is defined. It is the least positive integer with $a^k \\equiv 1 \\pmod p$.\n4. By [[Fermat's little theorem|0M25PRN3WQ]], $a^{p-1} \\equiv 1 \\pmod p$.\n5. By the [[division algorithm|0CHQKZFK5S]], there are integers $q$ and $r$ with $p - 1 = qk + r$ and $0 \\le r \u003c k$. Here $q \\ge 0$, because $p - 1 \\ge 0$ and $r \u003c k$.\n6. Then $a^{p-1} = (a^k)^q \\cdot a^r$. Since $a^k \\equiv 1$, [[powers|0J5VGJYYC3]] and [[products|072D5G4SK6]] of congruences give $a^{p-1} \\equiv 1^q \\cdot a^r = a^r \\pmod p$.\n7. With step 4 this gives $a^r \\equiv 1 \\pmod p$.\n8. If $r \u003e 0$, then $r$ is a positive integer less than $k$ with $a^r \\equiv 1 \\pmod p$. This contradicts the minimality of $k$.\n9. So $r = 0$, and $p - 1 = qk$. Hence $\\operatorname{ord}_p(a)$ divides $p - 1$.\n\n## Idea\n\nFermat's little theorem says that the exponent $p - 1$ brings $a$ back to $1$. Any exponent that does so is a multiple of the order, since the remainder on division by the order would be a smaller such exponent.\n",
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