The zeta function again

Last time we saw a fun identity, and used it to show that there were only finitely many primes. That identity was:

$\zeta(s)=\displaystyle\sum_{n=1}^\infty n^{-s}=\prod_p\frac1{1-p^{-s}}$

Our proof last time was that $\zeta(1)$ diverged, and so there had to be infinitely many primes. We’re going to notice something slightly different this time: A fun fact is that $\zeta(2)=1+\frac14+\frac19+\cdots=\frac{\pi^2}6$, an irrational number. Once again, I will not prove this here. Rest assured that to prove this, it is not necessary to know that there are infinitely many primes (so our proof isn’t circular).

But notice that for every prime $p$, $\frac1{1-p^{-2}}$ is always rational. No finite product rational numbers can be irrational, so there must be infinitely many primes!

This same technique works with any $\zeta(2k)$. It probably works for odd numbers too, but it’s unknown whether or not these are irrational.