Existence of holomorphic n-th root and non-vanishing

Question

Suppose $f \in H(\Omega)$, $\Omega =$ arbitrary region. Suppose $f$ has a holomorphic $n-$th root in $\Omega$ for every positive integer $n$. Then I need to show that $f(z)\neq 0$ for all $z \in \Omega$.

Answer 1

Let $z_0\in G$ and let $ord_{z_0}(f)\in \mathbb N$ be the order of vanishing of $f$ at $z_0$. If $f=g^n$ we have $ord_{z_0}(f)=n\cdot ord_{z_0}(g)$. Hence $ord_{z_0}(f)$ is being divisible by all integers must be zero; in other words $f$ does not vanish at $z_o$.

But this is small beer.

The much stronger conclusion of your hypothesis is that actually $f$ is an exponential: there exists $h\in H(\Omega)$ with $f=e^h$.
This is an old chestnut that you can find for example in Remmert's Theory of Complex Functions, Chapter 9.