I'm interested in providing a non-topological (i.e no covering theory) proof of the following fact:
Let $\Omega$ be an open connected subset of $\Bbb C$. Prove that $f\colon \Omega \to \Bbb C\setminus 0$ analytic has an analytic $n$-th root (i.e. an analytic function $g\colon \Omega \to \Bbb C\setminus 0$ s.t. $g^n=f$) if and only if $$ \dfrac{1}{2\pi i}\int_{\gamma} \dfrac{f'}{f}dz \in n\Bbb Z$$ for every loop $\gamma$ in $\Omega$.
the only if part is trivial, and I can provide a covering theory proof of the if part (basically one prove that $f$ lifts to the $n$-fold cover $\Bbb C\setminus 0 \xrightarrow{(\cdot)^n} \Bbb C \setminus 0$).
I would like to see a more "complex analysis" flavoured proof. I tried working with local $n$-th roots (on simply connected domains in $\Omega$ and try to prove whether they patch together but I can't see how to use the hypothesis.
Can someone give me any advice?
You probably know that if $f$ has a holomorphic logarithm, call it $g$, then $z \mapsto \exp \frac{g(z)}{n}$ is a holomorphic $n^{\text{th}}$ root of $f$. And in that case, we have
$$g'(z) = \frac{f'(z)}{f(z)}\,,$$
so the $n^{\text{th}}$ root of $f$ is given by
$$r(z_0)\cdot \exp \biggl(\frac{1}{n} \int_{\gamma_z} \frac{f'(z)}{f(z)}\,dz\biggr) \tag{1}$$
where $r(z_0)$ is an $n^{\text{th}}$ root of $f(z_0)$ and $\gamma_z$ is a (continuous, piecewise continuously differentiable) path from $z_0$ to $z$ in $\Omega$.
Now you can check that under the given hypotheses, $(1)$ is a well-defined holomorphic function on $\Omega$ that defines an $n^{\text{th}}$ root of $f$.