A simple way to obtain $\prod_{p\in\mathbb{P}}\frac{1}{1-p^{-s}}=\sum_{n=1}^{\infty}\frac{1}{n^s}$.

I absolutely love this result, I literally cannot stop from getting goosebumps and smiling whenever I think about it. It is a proof from probability theory! I learned it in David William's Probability with Martingales, of which it is part of exercise E4.2.

Fix $s>1$ and recall that $\zeta(s) = \sum_{n \in \mathbb{N}} n^{-s}$, so we aim to show that $1/\zeta(s) = \prod_p(1-p^{-s})$ where of course $p$ ranges over the primes.

First, define a probability measure $P$ and an $\mathbb{N}$-valued random variable $X$ such that $P(X=n) = n^{-s}/\zeta(s)$ (for example take $P(\{n\}) = n^{-s}/\zeta(s)$ and $X(\omega)=\omega$). Let $E_k := \{X \text{ is divisible by } k\}$. We claim that the events $(E_p : p \text{ prime})$ are independent. We note that $$ P(E_k) = \sum_{i=1}^\infty P(X=ik) = \sum_{i=1}^\infty \frac{(ik)^{-s}}{\zeta(s)} = k^{-s} \frac{\zeta(s)}{\zeta(s)} = k^{-s}. $$

Then if $p_1,\ldots,p_n$ are distinct primes we have $$\bigcap_{i=1}^n E_{p_i} = E_{\prod_{i=1}^np_i},$$ so that $$ P\left(\bigcap_{i=1}^n E_{p_i}\right) = P(E_{\prod_{i=1}^np_i}) = \left(\prod_{i=1}^n p_i \right)^{-s} = \prod_{i=1}^n p_i^{-s} = \prod_{i=1}^n P(E_{p_i}) $$ so our independence claim is proved. Then we note that $1$ is the unique positive integer which is not a multiple of any prime. Hence $$ \frac{1}{\zeta(s)} = P(X=1) = P\left(\bigcap_p E_p^c\right) = \prod_p(1-P(E_p)) = \prod_p(1-p^{-s}). $$

Let $s$ for which $\Re(s)>1$ then, for all $p \in \mathbb{P}$ we have $$\sum_{k=1}^{\infty} \frac{1}{p^{ks}}=\left(1-\frac{1}{p^s}\right)^{-1}$$

Now let $A(N)$ be the set of all strictly positive numbers such as all prime divisors are at maximum $N$.

Then $$\prod_{p \in \mathbb{P},\text{ }p<N}\left(1-\frac{1}{p^s}\right)^{-1}=\sum_{n \in A(N)}\frac{1}{n^s}$$

Obviously $\{1,...,N\}\subset A(N)$, then

$$\left|\zeta(s)-\prod_{p \in \mathbb{P},\text{ }p<N}\left(1-\frac{1}{p^s}\right)^{-1}\right|\le \sum_{n>N}\frac{1}{n^{\Re{(s)}}}$$

When $N$ goes to infinity it gives the result.