Validity of challenge to common Euler product derivation strategy (infinitely long denominator prime factorisations) Background
The following Euler product for the Riemann zeta function is well known.
$$ \sum_n \frac{1}{n^s} = \prod_p (1-\frac{1}{p^s})^{-1} $$
Here $n$ ranges over all integers, $p$ over a primes, and real $s>1$.

Common Proof Strategy
Many derivations / proofs, found in textbooks and papers, consider the following expression.
$$(1 - \frac{1}{p^s})^{-1} = 1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \frac{1}{p^{3s}} + \ldots$$
The LHS is finite for any given $p$ and the series expansion is valid because $\frac{1}{p} < 1$.
The following takes the product over all primes.
$$\prod_{p_i} (1-\frac{1}{p_i^s})^{-1} = 1 + \frac{1}{p_1^s} + \frac{1}{p_1^{2s}} + \ldots  + \frac{1}{p_1^sp_2^{s}} + \frac{1}{p_1^sp_3^{s}} + \ldots $$
The LHS is a product of finite and non-zero factors.
The RHS has terms of the form $\frac{1}{X}$ where $X$ contains all combinations of the primes, and all combinations of powers of the primes.
It is common to apply the Fundamental Theorem of Arithmetic to see that there is one term X for each integer $n$, and therefore the RHS is the desired $\sum\frac{1}{n^s}$.

Challenge
A challenge (for example here) that has been raised to this very common proof logic is that there are terms $X$ with an infinite number of factors in the denominator, for example:
$$ \frac{1}{(2^2\cdot3^2\cdot 5^2\cdot 7^2 \cdot\ldots)^s} $$
or another simpler example:
$$ \frac{1}{(2\cdot 2 \cdot 2\cdot 2\cdot\ldots)^s} $$

Question
Is the challenge valid?
I am not a trained mathematician, but in my opinion the proof strategy is valid because terms $X$ with denominators with an infinite number of prime factors are equivalent to zero. That is:
$$ \frac{1}{(2^2\cdot3^2\cdot 5^2\cdot 7^2 \cdot\ldots)^s} = 0$$
and
$$ \frac{1}{(2\cdot 2 \cdot 2\cdot 2\cdot\ldots)^s} = 0$$
My assertion is that the proof strategy remains valid because any finite integer $n$ has a single finite non-zero term $X$, and those $X$ with infinitely long denominators can be discarded because they are zero.
 A: The way to prove this rigorously is to show that:
$$\sum_{n\leq N}\frac1{n^s}\leq \prod_{p\leq N}\left(1-1/p^s\right)^{-1}\leq\sum_{n=1}^{\infty}\frac1{n^s}$$ This you can get because the product in the middle is finite, so you can use the argument without worrying about infinitely many primes.
Then use the squeeze theorem as $N\to\infty.$
This technique works only for real $s>1,$ because the inequalities are not true for $s$ complex.

At heart the proof is to let $$A_N=\{n\geq 1\mid n\text{ has no prime factors }>N\}$$
Then you can show that:
$$\prod_{p\leq N}\left(1-1/p^s\right)^{-1} =\sum_{n\in A_N} \frac1{n^s}$$
Then you get the left-hand inequality because $1,2,\dots,N\in A_N$ and the right hand side because $A_N\subseteq \mathbb Z^+.$
A: By definition, where $p_i$ is the $i^{th}$ prime,
$$\prod_{p_i} \left(1-\frac{1}{p_i^s}\right)^{-1}:=\prod_{i=1}^\infty\left(1-\frac{1}{p_i^s}\right)^{-1}:= \lim_N \prod_{i=1}^N\left(1-\frac{1}{p_i^s}\right)^{-1}$$
Now $$\prod_{i=1}^N\left(1-\frac{1}{p_i^s}\right)^{-1} = \prod_{i=1}^N\left(\sum_{e_i=0}^\infty \frac 1{\,p_i^{se_i}\,}\right)$$
And
$$\begin{align}\prod_{i=1}^N\left(\sum_{e_i=0}^\infty \frac 1{\,p_i^{se_i}\,}\right) &= \prod_{i=1}^N \lim_{n_i} \sum_{e_i=0}^{n_i} \frac 1{\,p_i^{se_i}\,}\\
&= \lim_{n_1,\dots,n_N} \sum_{e_1=0}^{n_1}\dots\sum_{e_N=0}^{n_N}\frac 1{(p_1^{e_1}p_2^{e_2}\dots p_n^{e_N})^s}\end{align}$$
provided the RHS converges. But since the terms are all positive, the limitend is an increasing function of all its indices. And it is the sum of a finite subsequence of the known convergent sequence $\sum_k \frac 1{k^s}$, so it is bounded above by that value. Hence the RHS side must converge.
Further, it includes a term of the form $\frac 1{k^s}$ for every $k < p_{N+1}$. Thus
$$\sum_{k=1}^{p_{N+1}-1} \dfrac1{k^s} \le \lim_{n_1,\dots,n_N} \sum_{e_1=0}^{n_1}\dots\sum_{e_N=0}^{n_N}\frac 1{(p_1^{e_1}p_2^{e_2}\dots p_n^{e_N})^s} \le \sum_{k=1}^\infty \dfrac1{k^s}$$
$$\sum_{k=1}^{p_{N+1}-1} \dfrac1{k^s} \le \prod_{i=1}^N\left(1-\frac{1}{p_i^s}\right)^{-1}\le \sum_{k=1}^\infty \dfrac1{k^s}$$
Since $p_{N+1} \to \infty$ as $N \to \infty$, by the squeeze theorem
$$\sum_{k=1}^\infty \dfrac1{k^s} \le \prod_{i=1}^\infty\left(1-\frac{1}{p_i^s}\right)^{-1} \le \sum_{k=1}^\infty \dfrac1{k^s}$$
That is, $$\prod_{p_i} \left(1-\frac{1}{p_i^s}\right)^{-1} = \sum_{k=1}^\infty \dfrac1{k^s}$$
Note that at one stage in this proof, it required that $$\sum_{k=1}^\infty \dfrac1{k^s}$$ converge. I.e., it only works for $s > 1$.
