From the discussion A simple way to obtain \prod_{p \in \mathbb{P}} \frac{1}{1-p^{-s}} = \sum_{n =1}^{\infty} \frac{1}{n^s} on math.stackexchange, I learnt an elementary proof of Euler product formula; in fact, I found it so remarkable that I decided to write a post on it.

We only need the two following basic lemmas; only sketchs of proof are given, we refer to Jean Jacod’s book, Probability essentials, for more information.

Definition: \{ E_i \mid i \in I \} is a family of independent events if for every finite subset J \subset I,

\displaystyle P \left( \bigcap\limits_{j \in J} E_j \right)= \prod\limits_{j \in J} P(E_j).

Lemma: Let P be a probability measure and \{ E_i \mid i \geq 1 \} be a family of independent events. Then

\displaystyle P \left( \bigcap\limits_{i \geq 1} E_i \right) = \prod\limits_{i \geq 1} P(E_i) .

Sketch of proof. \left( \bigcap\limits_{i=1}^n E_i \right) is a decreasing sequence of events, hence

\displaystyle P \left( \bigcap\limits_{i \geq 1} E_i \right) = \lim\limits_{n \to + \infty} P \left( \bigcap\limits_{i=1}^n E_i \right)= \lim\limits_{n \to + \infty} \prod\limits_{i=1}^n P(E_i) = \prod\limits_{i \geq 1} P(E_i). \square

Lemma: If \{ E_i \mid i \in I \} is a familiy of independent events, then \{ E_i^c \mid i \in I \} so is.

Sketch of proof. Using inclusion-exclusion principle, prove by induction on n that every subfamily \{E_{i_1}, \dots E_{i_n} \} of cardinality n satisfies

P \left( \bigcap\limits_{k=1}^n E_{i_k}^c \right)= \prod\limits_{k=1}^n P \left( E_{i_k}^c \right).

For n=2, we have

\begin{array}{lcl} P(E_1^c \cap E_2^c ) & = & 1- P(E_1 \cup E_2)= 1- P(E_1) - P(E_2 ) + P(E_1 \cap E_2) \\ \\ & = & 1- P(E_1) - P(E_2) + P(E_1) P(E_2) = ( 1 - P(E_1)) (1- P(E_2)) \\ \\ & = & P(E_1^c) P(E_2^c) \hspace{1cm} \square \end{array}

Theorem: Let s > 1. Then \displaystyle \zeta(s):= \sum\limits_{n \geq 1} \frac{1}{n^s} = \prod\limits_{p \in \mathbb{P}} \frac{1}{1-p^{-s}}.

Proof. Let X : \mathbb{N} \to \mathbb{N} be a random variable and P be a probability measure such that

\displaystyle P(X=n)= \frac{\zeta(s)^{-1}}{n^s};

for example, take X= \mathrm{Id} and P ( \{n \})= \frac{\zeta(s)^{-1}}{n^s}. Let E_k be the event “X is divisible by k“. Notice that

\displaystyle P(E_k)= \sum\limits_{i \geq 1} P(X=ik) = \sum\limits_{i \geq 1} \frac{\zeta(s)^{-1}}{n^s} = k^{-s}.

Therefore, for any primes k_1 , \dots, k_r,

\displaystyle P \left( \bigcap\limits_{j=1}^r E_{k_j} \right)= P \left( E_{\prod\limits_{j=1}^r k_j} \right) = \prod\limits_{j=1}^r k_j^{-s} = \prod\limits_{j=1}^r P(E_{k_j}).

We deduce that \{ E_k \mid k \ \text{prime} \} is a family of independent events. Because X=1 if and only if X is not divisible by any prime, we have

\displaystyle \frac{1}{\zeta(s)} = P(X=1)= P \left( \bigcap\limits_{p \in \mathbb{P}} E_p^c \right) =\prod\limits_{p \in \mathbb{P}} \left( 1- P(E_p) \right) = \prod\limits_{p \in \mathbb{P}} \left( 1-p^{-s} \right). \square