= Solution
The <Möbius function> has $\mu(1)=1$, $\mu(n)=0$ when $n$ is divisible by a <prime> square, and $\mu(n)=(-1)^r$ when $n$ is a product of $r$ distinct <primes>. The <Riemann zeta function> is initially the <Dirichlet series> $\zeta(s)=\sum_{n\ge1}n^{-s}$ for $\sigma=\Re s>1$. Both this <Dirichlet series> and $\sum\mu(n)n^{-s}$ have <absolute convergence>, because $|\mu(n)|\le1$ and $\sum n^{-\sigma}<\infty$.
For a finite set of <primes> $p\le y$, multiplication of absolutely convergent <geometric series> and <unique prime factorization> give
$$
\prod_{p\le y}(1-p^{-s})^{-1}=\sum_{\substack{n\ge1\\p\mid n\Rightarrow p\le y}}n^{-s}.
$$
Every omitted integer has a <prime factor> exceeding $y$, so it exceeds $y$. The difference from $\zeta(s)$ has absolute value at most $\sum_{n>y}n^{-\sigma}\to0$. This proves the <Euler product>, with the product interpreted as the limit of its finite products:
$$
\boxed{\zeta(s)=\prod_p(1-p^{-s})^{-1}\qquad(\Re s>1).}
$$
For the reciprocal identity, <absolute convergence> justifies multiplying and regrouping the two <Dirichlet series> by their product index. Their <Dirichlet convolution> coefficient at $m$ is
$$
\sum_{d\mid m}\mu(d)=\begin{cases}1,&m=1,\\(1-1)^r=0,&m>1\text{ with }r\text{ distinct prime factors}.
\end{cases}
$$
The coefficient calculation uses only the <squarefree integers> dividing $m$. Consequently \b[the reciprocal identity is an identity of absolutely convergent series]:
$$
\boxed{\zeta(s)\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=1\qquad(\Re s>1).}
$$
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