The Riemann zeta function has its Euler product on the half-plane $\mathrm{Re}s>1$

Write $\sigma :=\mathrm{Re}s$. Since the primes are among the integers at least $2$, ${\sum }_p\mathrm{\mid }{p}^{-s}\mathrm{\mid }={\sum }_p{p}^{-\sigma }\le {\sum }_{n=2}^{\mathrm{\infty }}{n}^{-\sigma }<\mathrm{\infty }$ by [L1]. Therefore [L2] applies to $a_p:=-p^{-s}$, so the product ${\prod }_p(1-p^{-s})$ converges and is nonzero.

For a finite set $P$ of primes, ${\prod }_{p\in P}\frac{1}{1-p^{-s}}={\prod }_{p\in P}{\sum }_{k\ge 0}{p}^{-ks}\mathrm{.}$ Multiplying out this finite product lists exactly the terms $n^{-s}$ for those integers $n\ge 1$ whose prime divisors all lie in $P$, and [L3] with [L4] shows that each such integer appears exactly once. Hence ${\prod }_{p\in P}\frac{1}{1-p^{-s}}={\sum }_{\begin{array}{c}n\ge 1\\ p\mid n\Rightarrow p\in P\end{array}}{n}^{-s}\mathrm{.}$

Let $P_N$ be the set of primes at most $N$. By step 1.2 the corresponding partial products are the partial sums over integers all of whose prime divisors lie in $P_N$. Every fixed integer eventually has this property, and the omitted terms are bounded in absolute value by the tail of the absolutely convergent series in [L1]. Therefore these partial products converge pointwise to ${\sum }_{n\ge 1}{n}^{-s}=\zeta (s)$.

Let $K\subseteq \{s:\mathrm{Re}s>1\}$ be compact, and choose $\sigma >1$ with $\mathrm{Re}s\ge \sigma$ on $K$. If an integer $n$ is omitted from the partial product over $P_N$, then $n$ has some prime divisor greater than $N$, hence $n>N$. So for $s\in K$, $\mid \zeta (s)-{\prod }_{p\le N}\frac{1}{1-p^{-s}}\mid \le {\sum }_{n>N}{n}^{-\sigma }\mathrm{.}$ The tail on the right tends to $0$ independently of $s$, so the partial products converge uniformly on $K$. Thus the Euler product converges locally uniformly on $\mathrm{Re}s>1$. Combining this with steps 1.1 and 2.1 proves the stated formula together with its absolute and locally uniform convergence.