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.
- Sieve proof
By sieving, we see that for every prime $q$ $$\left(\prod_{p\,\text{prime}\le q}\left(1-\frac{1}{p^s}\right)\right)\zeta (s)=\sum_{n;2,3,\ldots ,q\nshortmid n}\frac{1}{n^s},\quad \Re (s)\gt 1$$ 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 $$\begin{align}\left|\left(\prod_{p\,\text{prime}\le q}\left(1-\frac{1}{p^s}\right)\right)\zeta (s)-1\right|&=\left|\sum_{n;n\ne 1 \& 2,3,\ldots q,\nshortmid n}\frac{1}{n^s}\right|\\ &\le \sum_{n;n\ne 1 \& 2,3,\ldots q,\nshortmid n}\frac{1}{n^r}\\ &\le \sum_{n=q}^\infty \frac{1}{n^r},\quad \Re (s)\gt 1.\end{align}$$ 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 $$\left(\prod_{p\,\text{prime}}\left(1-\frac{1}{p^s}\right)\right)\zeta (s)=1,\quad \Re (s)\gt 1.$$
- 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.
This led me to the following question: Is there a proof of Euler's product for $\zeta (s)$ without using infinitude of primes? Note that this is not a duplicate of Proof of Infinitude of Primes by Euler's Product Formula is Circular? because the OP there fails to distinguish between the fundamental theorem of arithmetic and infinitude of primes.