### How the Riemann zeta function is connected to prime numbers

The Riemann zeta function is defined as
$$
\zeta(s) = \sum_{n=1}^{+\infty}\frac{1}{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots
$$

We can multiply each side by $(1-2^{-s})$
$$
(1-2^{-s})\zeta(s) = 
\left(\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots\right)
(1-2^{-s})
$$

and distribute
$$
(1-2^{-s})\zeta(s) = 
\left(\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots\right)
-
\left(\frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{6^s} + \cdots\right)
$$
to get
$$
(1-2^{-s})\zeta(s) = 
\left(\frac{1}{1^s} + \frac{1}{3^s} + \frac{1}{5^s} + \cdots\right)
$$

Continuing with this thought, we can multiply each side by $(1-3^{-s})$ to get rid of more terms
$$
(1-3^{-s})(1-2^{-s})\zeta(s) = 
\left(\frac{1}{1^s} + \frac{1}{5^s} + \frac{1}{7^s} + \frac{1}{11^s} + \cdots\right)
$$

We are just multiplying by $(1-p^{-s})$ where $p$ is prime,
 removing all terms whose denominators are divisible by $p^{s}$.
 If we keep going, there will be only one term left.
 
We can denote this as

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

or more simply

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