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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Let $n \u003e 1$ be an integer with $(n - 1)! \\equiv -1 \\pmod n$. Suppose that $n$ is not a [[prime number|0SN0MAEKYX]].\n2. Since $n \u003e 1$ is not prime, $n$ has a positive [[divisor|096PW7Q014]] $a$ other than $1$ and $n$. So $1 \u003c a \u003c n$.\n3. Since $1 \u003c a \\le n - 1$, the number $a$ is one of the factors of $(n - 1)! = 1 \\cdot 2 \\cdots (n - 1)$. So $a$ divides $(n - 1)!$.\n4. By the definition of [[congruence|0TPW25CBN6]], $n$ divides $(n - 1)! - (-1) = (n - 1)! + 1$. Since $a \\mid n$, also $a$ divides $(n - 1)! + 1$.\n5. By steps 3 and 4, $a$ divides $\\big( (n - 1)! + 1 \\big) - (n - 1)! = 1$.\n6. This is impossible, since $a \u003e 1$. So $n$ is prime.\n\n## Idea\n\nA proper divisor $a$ of $n$ appears as a factor in $(n - 1)!$. It also divides $(n - 1)! + 1$, so it would divide $1$, which cannot happen.\n",
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