TITLE: Alternative Proof of Infinitely Many Primes?
QUESTION [5 upvotes]: I've seen Euclid's proof of infinitely many primes, what are other approaches to proving there are infinitely many primes?

REPLY [6 votes]: Here's another proof.I thought of it a while ago, and posted it as an answer to a mathoverflow question, but I don't imagine I was the first to think of it. The idea is not to rely on the fact that $\pi$ is irrational. Let $\mathcal{P}$ denote the set of primes. We have 
$\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$ and $\sum_{n=1}^{\infty} \frac{1}{n^{4}} = \frac{\pi^{4}}{90}.$ The first equation yields, $\prod_{p \in \mathcal{P}}\frac{p^{2}}{p^{2}-1} = \frac{\pi^{2}}{6}$ and the second leads to $\prod_{p \in \mathcal{P}}\frac{p^{4}}{p^{4}-1} = \frac{\pi^{4}}{90}.$ If we square the first of the last two equations and divide by the second, we obtain $\prod_{p \in \mathcal{P}}\frac{p^{2}+1}{p^{2}-1} = \frac{5}{2}.$ If $\mathcal{P}$ was finite, the left hand expression would be a rational number whose numerator was not divisible by $3$, but whose denominator was divisible by $3$ (because $p^{2}+1 \equiv 2$ (mod $3$) for every prime other than $3$, and $p^{2}-1$ is divisible by $3$ for every prime other than $3$). However, no such rational number could be equal to $\frac{5}{2},$ a contradiction.