 Solution

For $\sigma =\mathrm{Re}s>1$, the Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ is absolutely convergent. A finite product over primes $p\le P$ can be expanded into absolutely convergent geometric series. By unique prime factorization,The omitted integers are all greater than $P$, so the difference from $\zeta (s)$ has modulus at most ${\sum }_{n>P}{n}^{-\sigma }\to 0$. ThusThis defines the infinite Euler product as the limit of finite products. Its absolute convergence also follows from ${\sum }_p{p}^{-\sigma }<\infty$, with $\log (1-p^{-s}{)}^{-1}=p^{-s}+O(p^{-2\sigma })$ for sufficiently large $p$.