Is this a way to prove there are infinitely many primes?

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?

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.