Proof of Euclid's theorem using n! + 1

Proof

We show that $n! + 1$ has a prime divisor larger than $n$.

1. Let $n$ be a natural number, and put $N = n! + 1$. Here $n! = 1 \cdot 2 \cdots n$, with $0! = 1$.
2. Since $n! \ge 1$, $N \ge 2$. By the existence of a prime divisor, there is a prime number $p$ with $p \mid N$.
3. Suppose $p \le n$. Since $p \ge 2$, $p$ is one of the factors $1, 2, \dots, n$ of $n!$, so $p$ divides $n!$.
4. Then $p$ divides $N - n! = 1$. But $p \ge 2$, and $2$ or more cannot divide $1$. This is a contradiction.
5. So $p > n$, and in particular $p \ge n$. This proves the second form of the statement.
6. Suppose there were only finitely many primes. Let $n$ be one more than the largest of them.
7. 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.