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! 
 A: 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.
