I was looking at Euler's product form for the Riemann zeta function, which I think is very cool. I thought I would have a go at proving it before looking at any proofs online. My training is in mathematical physics, I didn't take any courses on analysis in my undergrad and I'm trying to learn some.
Here is my proof;
Replace each factor in the product using the geometric series, $$\prod_{p \mathrm{prime}}\frac{1}{1-p^{-s}} = \prod_{p \mathrm{prime}}\left( \sum_{r_p=0}^\infty( p^{-s})^{r_p}\right).$$ Note that $p^{-s} < 1$ for all $p$ and all $s>0$, so we know that each series converges.
Then distribute the product over the sums $$ \prod_{p \mathrm{prime}}\left( \sum_{r_p=0}^\infty( p^{-s})^{r_p}\right)= \lim_{q\rightarrow\infty}\sum_{r\in\mathbb{N}^q} \prod_{p \mathrm{prime}<q} ( p^{-s})^{r_p}.$$
Now rearrange using power laws, $$\lim_{q\rightarrow\infty}\sum_{r\in\mathbb{N}^q} \prod_{p \mathrm{prime}<q} ( p^{-s})^{r_p} = \lim_{q\rightarrow\infty}\sum_{r\in\mathbb{N}^q} \left(\prod_{p \mathrm{prime}<q} p^{r_p} \right)^{-s}.$$
Finally note that by the fundamental theorem of arithmetic this sum goes over all natural numbers written in terms of their prime factorisation, and hence $$\lim_{q\rightarrow\infty}\sum_{r\in\mathbb{N}^q} \left(\prod_{p \mathrm{prime} <q} p^{r_p} \right)^{-s}=\sum_{n=1}^{\infty}\frac{1}{n^s}.$$
So I wanted to know, what would I need to justify for this proof to be rigours? It seems to me that the geometric series, power law rearrangement and fundamental theorem of arithmetic steps are all clear. The distributive step works for a finite product of finite sums, and it's intuitive to me that this would work for this infinite product of infinite sums, although this would require taking infinitely many limits. Is there any problem with doing this?
There is a Wikipedia page on proving that these two forms of the Riemann zeta function are equivalent. The first proof due to Euler using the sieve of Eratosthenes is clearly inspired and beautiful.
The second proof is similar to my proof, but it follows a slightly different reasoning which bounds the difference between the two forms for the zeta function for a finite number of primes. This proof is more similar to the kind of arguments for proofs involving sequences and series that I have seen when looking through introductory courses on analysis. Is there some reason why proofs like this seem to be preferred?
The proof I've given appears to me to be more direct because it only rearranges the expression at each step, in a kind of roughly algorithmic way. I wouldn't have thought to myself to try and bound the absolute value for the difference for a finite number of factors, this requires introducing something new so seems less intuitive to me. Maybe this is because I have not spent enough time reading analysis courses yet.