Let $f$ be analytic and nowhere zero on $0<|z|<1$. Prove that $f(z)=z^n \exp(g(z))$ for some integer $n$ and $g$ analytic in $0<|z|<1$.
$\ln f$ is defined locally on $0<|z|<1$ as $\ln f = \ln |f| + i \arg f$ for some branch of $\arg f$. The issue is whether or not we have a single valued branch of $\ln f$ in the annulus $0<|z|<1$. We must check that analytic continuation along every (simple) closed path comes back to the original branch. Any such path is homotopic to either a constant path (nothing to worry about) or to a circle about the origin. Analytic continuation along such a circle yields the original branch plus $+2\pi i n$ for some $n\in\mathbb Z$, due to the aforementioned connection with argument. If $n\ne 0$ we are in trouble. To get out of trouble, apply the above not to $f$ itself, but to its product with an appropriate power of $z$.