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  "body": "\n## Proof\n\n1. Let $p$ be a [[prime number|0SN0MAEKYX]]. If $p = 2$, then $(p - 1)! = 1$, and $2$ [[divides|096PW7Q014]] $1 - (-1) = 2$. So $(p - 1)! \\equiv -1 \\pmod p$. From now on let $p \\ge 3$.\n2. Let $R = \\{ 1, 2, \\dots, p - 1 \\}$. Each $a \\in R$ satisfies $p \\nmid a$, so the [[greatest common divisor|02WF878BH1]] $\\gcd(a, p)$ is a positive divisor of $p$ other than $p$. Thus $\\gcd(a, p) = 1$.\n3. Let $a \\in R$. By step 2 and the [[existence of inverses modulo n|0M67FWZWPQ]], $a$ has a [[modular multiplicative inverse|00GXTQMY8Z]] $x$, so $a x \\equiv 1 \\pmod p$.\n4. By the [[division algorithm|0CHQKZFK5S]], $x = q p + a'$ with $0 \\le a' \u003c p$. Then $a a' \\equiv a x \\equiv 1 \\pmod p$. Since $a \\cdot 0 = 0 \\not\\equiv 1 \\pmod p$, we get $a' \\ne 0$, so $a' \\in R$.\n5. If $b \\in R$ and $a b \\equiv 1 \\pmod p$, then $a b \\equiv a a' \\pmod p$. By [[cancellation|09RDZZGJZ9]] with $\\gcd(a, p) = 1$, $b \\equiv a' \\pmod p$. Both lie in $R$, so $b = a'$.\n6. So each $a \\in R$ has exactly one $a' \\in R$ with $a a' \\equiv 1 \\pmod p$. Since $a' a \\equiv 1 \\pmod p$, step 5 applied to $a'$ gives $(a')' = a$.\n7. Let $a \\in R$. Then $a' = a$ if and only if $a^2 \\equiv 1 \\pmod p$, that is, $p \\mid (a - 1)(a + 1)$.\n8. By [[Euclid's lemma|051VYKA1EX]], this holds if and only if $p \\mid a - 1$ or $p \\mid a + 1$. Since $0 \\le a - 1 \\le p - 2$ and $2 \\le a + 1 \\le p$, this means $a = 1$ or $a = p - 1$.\n9. Let $T = \\{ 2, 3, \\dots, p - 2 \\}$, which is empty when $p = 3$. Let $a \\in T$. If $a' = 1$, then $a \\equiv a a' \\equiv 1 \\pmod p$, so $a = 1$. If $a' = p - 1$, then $-a \\equiv a a' \\equiv 1$, so $a = p - 1$. Both contradict $a \\in T$.\n10. So $a \\mapsto a'$ maps $T$ to $T$. By steps 6 and 8, it is its own inverse and has no fixed point in $T$.\n11. Hence $T$ splits into disjoint pairs $\\{ a, a' \\}$ with $a \\ne a'$. Each pair has product $a a' \\equiv 1 \\pmod p$.\n12. By [[compatibility with multiplication|072D5G4SK6]], $(p - 2)! = 1 \\cdot \\prod_{a \\in T} a \\equiv 1 \\pmod p$. This also holds when $T$ is empty.\n13. Then $(p - 1)! = (p - 1) \\cdot (p - 2)! \\equiv (-1) \\cdot 1 = -1 \\pmod p$, by compatibility with multiplication and $p - 1 \\equiv -1 \\pmod p$.\n\n## Idea\n\nModulo a prime, every nonzero residue has a unique inverse. Only $1$ and $-1$ are their own inverses, so all the other factors of $(p - 1)!$ cancel in pairs. What remains is $1 \\cdot (p - 1) \\equiv -1$.\n",
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