TITLE: Is this a way to prove there are infinitely many primes?
QUESTION [21 upvotes]: Someone gave me the following fun proof of the fact there are infinitely many primes. I wonder if this is valid, if it should be formalized more or if there is a falsehood in this proof that has to do with "fiddling around with divergent series".
Consider the product 
$\begin{align}\prod_{p\text{ prime}} \frac p{p-1} &= \prod_{p\text{ prime}} \frac 1{1-\frac1p}\\&=\prod_{p\text{ prime}}\left(\sum_{i=0}^\infty\frac1{p^i}\right)\\&=(1+\tfrac12+\tfrac14+\ldots)(1+\tfrac13+\tfrac19+\ldots)\ldots&(a)\\&=\sum_{i=1}^\infty\frac1i&(b)\\&=\infty\end{align}\\\text{So there are infinitely many primes }\blacksquare$
Especially the step $(a)$ to $(b)$ is quite nice (and can be seen by considering the unique prime factorization of each natural number). Is this however a valid step, or should we be more careful, since we're dealing with infinite, diverging series?

REPLY [19 votes]: The rigorous approach, as noted in comments.
If there are only finitely many primes, pick an $n$ so that:
$$H_n=\sum_{m=1}^{n}\frac{1}m > \prod_{p}\frac{1}{1-\frac 1p}$$
You can do this because the right side is a finite product of positive real numbers, and the series $\sum_{m=1}^{\infty}\frac1m$ diverges.
Next, show that:
$$\prod_p \frac{1}{1-\frac{1}{p}} > \prod_p \sum_{k=0}^{\lfloor \log_{p}n\rfloor} \frac{1}{p^k}>\sum_{m=1}^{n}\frac{1}{m}$$
Reaching a contradiction.