I read in a text that Euler once proved the infinitude of primes by proving the divergence of their reciprocals, which seems to me a highly pleasing. However, I am only familiar with one proof of the primes' reciprocals' divergence and it requires the infinitude of the primes in one of its steps (it essentially compares the sum of primes cleverly with the harmonic series using some nice inequality manipulations).
I don't need to see Euler's original proof (which I am informed was insufficiently rigorous for my, modern, tastes anyway) but is there an elementary proof of the divergence of primes' reciprocals that does not require the infinitude of the primes?
Euler said.
$\zeta(s) = 1 + 2^{-s} + 3^{-s}+4^{-s} + \cdots\\ 2^{-s}\zeta(s) = 2^{-s} + 4^{-s} + 6^{-s} \cdots\\ (1-2^{-s})\zeta(s) = 1 + 3^{-s}+5^{-s} + \cdots\\ (1-2^{-s})(1-3^{-s})\zeta(s) = 1 + 5^{-s}+7^{-s} + 11^{-s} +\cdots$
This is effectively the sieve of Eranthoses
$\prod_\limits{p \text { prime}} (1-p^{-s})\zeta(s) = 1\\ \zeta(s) = \prod_\limits{p \text { prime}} \frac {p^s}{p^s-1}\\ \zeta(1) = 1 + \frac 12 + \frac 13 + \frac 14 + \cdots = \infty\\ \prod_\limits{p \text { prime}} \frac {p}{p-1} = \infty$
If the set of prime numbers were finite, then the left hand side would be finite, but as it isn't it ain't