## Introduction:

The Euler product formula states that if $\zeta(s)$ is the Riemann zeta function and $p$ is prime:

$$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p} \frac{1}{1-p^{-s}}$$

holds for all $s \in \mathbb{C}$ such that $\zeta(s)$ is absolutely convergent.

## Proof:

Every positive integer $n \in \mathbb{N^*}$ has a unique prime factorization:

$$\forall n \in \mathbb{N^*} \exists c_p \in \mathbb{N}, n = \prod_p p^{c_p}$$

where $% $.

Furthermore, we note that:

$$\prod_p \frac{1}{1-p^{-s}} = \prod_p \big(\sum_{c_p = 0}^\infty p^{-c_p s} \big)$$

due to elementary properties of geometric series.

In the formal expansion (2) we note that each term has a unique prime factorization and that every possible prime factorization occurs once. It follows that if $\sum n^{-s}$ converges absolutely we may rearrange the sum however we wish and so:

$$\zeta(s) = \prod_{p} \frac{1}{1-p^{-s}}$$

provided that the hypotheses on $s$ are satisfied.