Euler product formula

Statement

Let $s$ be a complex number with real part $\operatorname{Re}(s) > 1$, and let $\zeta(s) = \sum_{n=1}^{\infty} n^{-s}$ be the Riemann zeta function. Then $$\zeta(s) = \prod_{p} \frac{1}{1 - p^{-s}},$$ where the product runs over all prime numbers $p$ and converges absolutely.

Remarks

The proof expands each factor as a geometric series $1 + p^{-s} + p^{-2s} + \cdots$. The fundamental theorem of arithmetic shows that the product of these series contains each term $n^{-s}$ exactly one time. The same method gives a product formula $\sum_n f(n) n^{-s} = \prod_p \sum_{k \ge 0} f(p^k) p^{-ks}$ for each multiplicative function $f$ when the series converges absolutely. Since $\zeta(s) \to \infty$ as $s \to 1^+$, the formula gives a proof of Euclid's theorem.

Explanations

L1 human

Multiply $(1 + \tfrac12 + \tfrac14 + \cdots)(1 + \tfrac13 + \tfrac19 + \cdots)(1 + \tfrac15 + \cdots) \cdots$ and expand. Each fraction $1/n$ appears one time, because $n$ is a product of prime powers in exactly one way. The formula with the exponent $s$ is the same idea with $1 / n^s$.