There are two proofs of
$$\sum\_{n=1}^\infty \frac{1}{n^s}=\prod\_{p \text{ prime}}\frac{1}{1-p^{-s}},\quad \Re (s)\gt 1$$
which I'm aware of. I'll call the first one the **Sieve proof** and the second one will be the **Factorization proof**. Both of them use infinitude of primes (at least I think so).


1. Sieve proof  

  

By sieving, we see that for every prime $q$
$$\left(\prod\_{p\,\text{prime}\le q}\left(1-\frac{1}{p^s} \right) \right) \zeta (s)=\sum\_{n;2,3,\ldots ,q\,\nmid\, n}\frac{1}{n^s},\quad \Re (s)\gt 1$$
where the sum is over all $n\in\mathbb{N}$ that are not divisible by the primes from $2$ to $q$. Choosing $r\gt 1$ such that $|s|\gt r$ gives
$$\begin{align}\left|\left(\prod\_{p \text{ prime}\le q}\left(1-\frac{1}{p^s}\right)\right)\zeta (s)-1\right|&=\left|\sum\_{n;n\ne 1 \& 2,3,\ldots q,\,\nmid\, n}\frac{1}{n^s}\right|\\
&\le \sum\_{n;n\ne 1 \& 2,3,\ldots q,\\,\nmid\, n}\frac{1}{n^r}\\
&\le \sum\_{n=q}^\infty \frac{1}{n^r},\quad \Re (s)\gt 1.\end{align}$$
**Because of infinitude of primes**, we can let $q\to\infty$, and by Cauchy's criterion for series $\lim\_{q\to\infty}\sum\_{n=q}^\infty \frac{1}{n^r}=0$, so
$$\left(\prod\_{p\,\text{prime}}\left(1-\frac{1}{p^s}\right)\right)\zeta (s)=1,\quad \Re (s)\gt 1.$$
2. Factorization proof (this proof is taken from *The Theory of the Riemann Zeta-function* by Titchmarsh)  

  

We have
$$\prod\_{p\le P}\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\cdots\right)=1+\frac{1}{n\_1^s}+\frac{1}{n\_2^s}+\cdots$$
where $n\_1,n\_2,\ldots$ are those integers none of whose prime factors exceed $P$. It follows that
$$\begin{align}\left|\zeta (s)-\prod\_{p\le P}\left(1-\frac{1}{p^s}\right)^{-1}\right|&=\left|\zeta (s)-1-\frac{1}{n\_1^s}-\frac{1}{n\_2^s}-\cdots\right|\\
&\le \frac{1}{(P+1)^{\Re (s)}}+\frac{1}{(P+2)^{\Re (s)}}+\cdots\end{align}$$
This tends to $0$ as $P\to\infty$ (again **using infinitude of primes**) if $\Re (s)\gt 1$ and the product formula follows.


In [their comment](https://math.stackexchange.com/questions/4674095/proof-of-eulers-product-for-zeta-s-without-using-infinitude-of-primes#comment9880111_4674095) to [this question](https://math.stackexchange.com/questions/4674095/proof-of-eulers-product-for-zeta-s-without-using-infinitude-of-primes) the user J.G. commented that the proofs actually don't use infinitude of primes (which he didn't explain – so I'm very confused and I'm writing this question) and his comments contradict a textbook proof by Amann and Escher (see a [related question here](https://math.stackexchange.com/questions/4165001/zeta-function-product-in-amann-and-escher)) and a [Wikipedia proof](https://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function#Another_proof)  – notice "up to some **prime number limit** $q$".


This is not a duplicate of [this question](https://math.stackexchange.com/questions/3746532/proof-of-infinitude-of-primes-by-eulers-product-formula-is-circular) because the OP there fails to distinguish between the fundamental theorem of arithmetic and infinitude of primes.


No one answered on MSE so I'm posting here.



Do all rigorous proofs of Euler's product for $\zeta (s)$ use infinitude of primes?
Neither proof uses the infinitude of primes. Here is why ($p$ will always denote a prime number).


1. Sieve proof.


Proceed as you indicated, but don't assume that $q$ is prime. We still have that
$$\left|\left(\prod\_{p\le q}\left(1-\frac{1}{p^s}\right)\right)\zeta (s)-1\right|=\left|\sum\_{\substack{n>1\\\forall p:\ p\mid n\ \Rightarrow\ p>q}}\frac{1}{n^s}\right|
\le \sum\_{n>q}\frac{1}{n^{\Re(s)}},\qquad \Re (s)\gt 1.$$
The right-hand side tends to zero as $q\to\infty$, done.


2. Factorization proof


Proceed as you indicated, and recall that
$$\left|\zeta (s)-\prod\_{p\le P}\left(1-\frac{1}{p^s}\right)^{-1}\right|
=\left|\sum\_{\exists p:\ p\mid n\ \&\ p>P}\frac{1}{n^s}\right|
\le \sum\_{n>P}\frac{1}{n^{\Re(s)}},\qquad \Re (s)\gt 1.$$
The right-hand side tends to zero as $P\to\infty$, done.


