A simple proof of Euler's product formula

Introduction:

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

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

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:

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

where $\sum c_p < \infty$.

Furthermore, we note that:

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

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:

\begin{equation} \zeta(s) = \prod_{p} \frac{1}{1-p^{-s}} \end{equation}

provided that the hypotheses on $s$ are satisfied.