My question is about this problem:
Let $s > 1$ and $$\zeta(s) = \sum_{n = 1}^{\infty} n^{-s}.$$ Furthermore let $(\Omega, \mathcal{F}, \mathbb{P}$) be a probability space with $\Omega = \mathbb{N}$, $\mathcal{F} = 2^{\Omega}$ and $$\mathbb{P}({n}) = \frac{n^{-s}}{\zeta(s)}, n \in \mathbb{N}.$$
We define $\mathcal{P}_n = p$ is a prime number and $p \le n$ and $\mathcal{P}_\infty = \cup_{n \in \mathbb{N}} \mathcal{P}_n$.
Prove the following : $\mathbb{P}(n \in \mathbb{N} : k^2$ does not divide $n$ for all $k \ge 2$) means $$\prod_{p \in \mathcal{P}_\infty} (1 - p^{-2s})$$
I do not really how to start this, I am unable to find any link between the right-hand side and the left-hande of the equation. I am grateful for any tip, suggestion and piece of advice