Euclid's proof of the infinitude of primes

Suppose, for contradiction, that there are only finitely many primes, and let $p_1, p_2, \dots, p_n$ be the complete list of all of them.

Consider the number $N = p_1 p_2 \cdots p_n + 1$. Since $N > 1$, it must have at least one prime divisor; call it $p$ (every integer greater than $1$ has a prime factor).

By assumption, $p_1, \dots, p_n$ is the list of all primes, so $p$ must equal $p_i$ for some $i$. In particular, $p$ divides the product $p_1 p_2 \cdots p_n$.

But $p$ also divides $N = p_1 p_2 \cdots p_n + 1$. If $p$ divides both $p_1 p_2 \cdots p_n$ and $N$, then $p$ divides their difference: $p \mid \big(N - p_1 p_2 \cdots p_n\big) = 1$.

No prime number can divide $1$, since every prime is greater than $1$. This is a contradiction. Therefore the original assumption — that there are only finitely many primes — must be false, so there are infinitely many primes.