Question on a proof of the Euler product of zeta function

Let $\zeta(s)$ be the Riemann zeta function. Then we know it satisfies the Euler product for $\text{Re}(s) > 1$, $$\zeta(s) = \prod_{p} (1 - p^{-s})^{-1}.$$

The proof I read, if I recall correctly, was by considering the partial product of the right hand side, bound $\zeta(s)$ from above and below and take the limit, to prove this statement.

Then someone told me that since each one of $1/n^s$ appears precisely once on the right hand side, so we have the above equality, and that's all we need to do. This is intuitively clear to me, but I was wondering if this was a 'rigorous' proof of the Euler product of zeta function or not. I would greatly appreciate some clarification. Thank you very much!

• The key is to show the sum total of all terms of the product that have more than finitely many $1$'s contributes zero to the sum. That proof is in most elementary number theory books, have you looked there? Feb 9, 2016 at 17:35
• @GregoryGrant Thank you for your comment. I have seen proofs of this theorem before. What I was wondering was that if what this person told me is a rigorous proof or not, basically claiming that the proof is one line... Feb 11, 2016 at 17:28
• The manipulation of geometric series, and invocation of unique factorization (as in @Maurice's answer) is absolutely the idea, and should persuade anyone of the truth of the assertion. However, it is not quite completely rigorous, since some rationalization is needed to justify rearranging infinite expressions, akin to Fubini-Tonelli. But that's just a technicality, surely. Jul 7, 2017 at 18:03
• In this answer a completely formal proof is given along with a heuristic proof. In the formal proof the sieving out method is formalized by showing all the steps involved in manipulating the infinite series using 'sub-series'. Then the general relation (equation 5), in which the first $k$ primes are sieved out, is proved by induction. The equation (5) is then rearranged to derive an inequality showing formally the convergence of the infinite product to $\zeta(s)$ as $k \rightarrow \infty$. Nov 10, 2022 at 18:48

The proof is rigorous but takes more than one line for those less facile with mental mathematics. In the first step, we use the fact that $(1 - p^{-s})^{-1}$ is the sum of a geometric series with first term $1$ and common ratio $1/p^s$, Re$(s) > 1$. In the third step, every number of the form $\prod_p p^\alpha$, where $\alpha$ is a non-negative integer, occurs in the denominator of a term precisely once and is raised to the power $s$. Hence, by the fundamental theorem of arithmetic, each positive integer raised to the power $s$ occurs precisely once in the denominator of a term, which explains the fourth step. \begin{align} \prod_p ( 1 - p^{-s})^{-1} & = \prod_p \left(1 + \frac1{p^s} + \frac1{p^{2s}} + \cdots \right)\\ & = \left(1 + \frac1{2^s} + \frac1{2^{2s}} + \cdots \right) \left(1 + \frac1{3^s} + \frac1{3^{2s}} + \cdots \right) \left(1 + \frac1{5^s} + \frac1{5^{2s}} + \cdots \right) \cdots\\ & = 1 + \frac1{2^s} + \frac1{3^s} + \frac1{2^{2s}} + \frac1{5^s} + \frac1{2^s3^s} + \cdots\\ & = \frac1{1^s} + \frac1{2^s} + \frac1{3^s} + \frac1{4^s} + \frac1{5^s} + \frac1{6^s} + \cdots\\ & = \sum_{n = 1}^\infty \frac1{n^s}\\ & = \zeta(s) \end{align} Here is the justification for manipulating the factors and terms in the above proof: Because $\sum_p 1/p^s$, Re$(s) > 1$, converges absolutely, we can rearrange the factors in the first two lines above without affecting the product as explained here. Because $\sum_{k=0}^\infty 1/p^{ks}$, Re$(s) > 1$, converges absolutely, we can rearrange the terms in each of the factors in the second line without affecting the sum as explained here. Finally, because the product of absolutely convergent series is an absolutely convergent series, the series in the third line is absolutely convergent, so we can rearrange the terms in the third and fourth lines without affecting the sum.