<s>[INST] Euler product formula Why is it that
$$\prod_{p\,\le\, y}\sum_{k=0}^\infty p^{-ks} = \sum_{n\,\in\,\{n \,:\, p \,\mid\, n \, \Rightarrow \, p\,\le\,y\}} n^{-s} \text{ ?}$$
I understand that we have a product of finitely many absolute convergent series and therefore rearrangement is allowed, but I do not see how exactly the terms are being rearranged.
 [/INST] Your product $\displaystyle\prod_{p\,:\,p\,\le\,y}$ specifies $p\le y$, so let's try $y=5$, so that $p\in\{2,3,5\}.$ Your sum $\displaystyle \sum_{n \, : \, p \, \mid \, n,\ p\,\le\,y}$ requires $n$ to be divisible only by those primes, so where looking at integers divisible only by $2$, $3$, or $5$ and by no other primes.
\begin{align}
& \left( 1 + \frac 1 {2^s} + \frac 1 {4^s} + \frac 1 {8^s} + \cdots \right) \\
{}\times {} & \left( 1 + \frac 1 {3^s} + \frac 1 {9^s} + \frac 1 {27^s} + \cdots \right) \\
{}\times{} & \left( 1 + \frac 1 {5^s} + \frac 1 {25^s} + \frac 1 {125^s} + \cdots \right) \\[10pt]
= {} & \text{a sum in which, for example, } \frac 1 {7200^s} \text{ appears because} \\[5pt]
& 7200 = 2^5 \times3^3\times5 \text{ has no prime} \\[5pt]
& \text{factors besides 2, 3, and 5;} \\[5pt]
& \text{thus } \frac 1 {7200^s} = \frac 1 {2^{5s}} \times \frac 1 {3^{2s}} \times \frac 1 {5^s}. \tag a
\end{align}
Within the expression $\displaystyle \left( 1 + \frac 1 {2^s} + \frac 1 {4^s} + \frac 1 {8^s} + \cdots \right)$ one finds the term $1/2^{5s}$; specifically, it is the sixth term in that sum. Likewise, within the sum $\displaystyle \left( 1 + \frac 1 {3^s} + \frac 1 {9^s} + \frac 1 {27^s} + \cdots \right)$ one finds $1/3^{2s}$; specifically, it is $1/9^s$. And within $\displaystyle \left( 1 + \frac 1 {5^s} + \frac 1 {25^s} + \frac 1 {125^s} + \cdots \right)$ one finds $1/5^s$. Hence the term in the line labeled $(\text{a})$ above appears.  And so do all other numbers $1/n^s$ where $n$ is a number divisible only by primes that are $\le y$.
It is necessary to rely on the fact that each number has only one prime factorization: If there were two prime factorizations of $7200$, then we would see $\dfrac 2 {7200^s}$ instead of $\dfrac 1 {7200^s}.$
</s>