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  "body": "\n## Proof\n\nWe show that $n! + 1$ has a prime divisor larger than $n$.\n\n1. Let $n$ be a natural number, and put $N = n! + 1$. Here $n! = 1 \\cdot 2 \\cdots n$, with $0! = 1$.\n2. Since $n! \\ge 1$, $N \\ge 2$. By the [[existence of a prime divisor|0SPBEDVNW1]], there is a [[prime number|0SN0MAEKYX]] $p$ with $p \\mid N$.\n3. Suppose $p \\le n$. Since $p \\ge 2$, $p$ is one of the factors $1, 2, \\dots, n$ of $n!$, so $p$ [[divides|096PW7Q014]] $n!$.\n4. Then $p$ divides $N - n! = 1$. But $p \\ge 2$, and $2$ or more cannot divide $1$. This is a contradiction.\n5. So $p \u003e n$, and in particular $p \\ge n$. This proves the second form of the statement.\n6. Suppose there were only finitely many primes. Let $n$ be one more than the largest of them.\n7. By step 5, there is a prime $p \\ge n$. Then $p$ is larger than every prime, which is impossible. So there are infinitely many primes.\n\n## Idea\n\nEvery number from $2$ to $n$ divides $n!$, so it leaves remainder $1$ when it divides $n! + 1$. Hence every prime factor of $n! + 1$ is larger than $n$, and there is always a prime beyond any given bound.\n",
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