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