= Solution
Define the <Möbius function> by $\mu(1)=1$, $\mu(n)=(-1)^r$ for a product of $r$ distinct primes, and $\mu(n)=0$ if a prime square divides $n$. For $n>1$, summing over square-free divisors gives $\sum_{d\mid n}\mu(d)=(1-1)^r=0$, while the sum is one at $n=1$.
Both <Dirichlet series> $\sum n^{-s}$ and $\sum\mu(n)n^{-s}$ converge absolutely for $\sigma=\operatorname{Re}s>1$. Their product may therefore be regrouped by the product of the indices:
$$
\zeta(s)\sum_{n=1}^\infty\frac{\mu(n)}{n^s}
=\sum_{n=1}^\infty\frac{\sum_{d\mid n}\mu(d)}{n^s}=1.
$$
This simultaneously proves $\boxed{\zeta(s)\ne0\text{ for }\sigma>1}$ and
$$
\boxed{\frac1{\zeta(s)}=\sum_{n=1}^\infty\frac{\mu(n)}{n^s}.}
$$
In particular $|1/\zeta(\sigma+it)|\le\sum|\mu(n)|n^{-\sigma}\le\zeta(\sigma)$.
Under the assumed analytic continuation, $\zeta$ is holomorphic near $s_0=1+it_0$ when $t_0\ne0$, and $\zeta(\sigma)=1/(\sigma-1)+O(1)$ as $\sigma\downarrow1$. If $s_0$ were a zero of order $m\ge2$, local factorization would give $|1/\zeta(s_0+h)|\ge c h^{-m}$ for sufficiently small positive $h$. The preceding absolutely convergent bound gives instead $|1/\zeta(1+h+it_0)|=O(h^{-1})$, a contradiction. Thus $\boxed{m\le1}$. This <reciprocal Dirichlet-series bound on boundary-zero multiplicity> does not assert that any such boundary zero exists.
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