= Solution
For $\sigma=\operatorname{Re}s>1$, the <Dirichlet series> $\zeta(s)=\sum_{n\geq1}n^{-s}$ is absolutely convergent. A finite product over primes $p\leq P$ can be expanded into absolutely convergent <geometric series>. By <unique prime factorization>,
$$
\prod_{p\leq P}(1-p^{-s})^{-1}=\sum_{\substack{n\geq1\\\text{all prime factors of }n\leq P}}n^{-s}.
$$
The omitted integers are all greater than $P$, so the difference from $\zeta(s)$ has modulus at most $\sum_{n>P}n^{-\sigma}\to0$. Thus
$$
\boxed{\zeta(s)=\prod_p(1-p^{-s})^{-1}\qquad(\operatorname{Re}s>1).}
$$
This 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 $p$.
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