Suppose $\Omega$ is a domain of the complex plane (i.e. an open and connected subset of the plane). Suppose $f$ is holomorphic on $\Omega$, and $f$ is not identically zero.
Suppose $f$ has a holomorphic logarithm on $\Omega$, which means that there is a function $g$ holomorphic on $\Omega$ such that $e^g=f$. Then it is easy to show that $f$ has holomorphic $n$-th roots on $\Omega$ for each $n$, which means that for each integer $n$, there exist a function $g_n$ holomorphic on $\Omega$ such that $(g_n)^n = f$.
Is the converse true? i.e. if $f$ has holomorphic $n$-th roots on $\Omega$ for all $n$, then can we find a function $g$ holomorphic on $\Omega$ such that $f=e^g$?
A few remarks :
One can prove that if $f$ has holomorphic $n$-th roots on $\Omega$ for all $n$, then $f$ does not vanish on $\Omega$. Therefore, we can define a holomorphic logarithm locally, but is it possible to find a global holomorphic logarithm?
Furthermore, notice that $\Omega$ is not supposed simply connected, in which case the answer to my question is yes.
The condition that $f$ have a holomorphic logarithm is equivalent to
$df/f=f'(z)dz/f(z)$ being an exact differential. This is equivalent to
the integral of $df/f$ over all closed curves in $\Omega$ vanishing.
Let $C$ be a closed curve in $\Omega$.
If $f=g^n$ is an $n$-th power in $\Omega$ of a holomorphic $g$ then
$\int_C df/f=n\int_C dg/g$. But $\int_C dg/g$ is an integer multiple of $2\pi i$.
Hence $\int_C df/f$ is an integer multiple of $2\pi ni$. If this holds for
all $n$ then $\int_C df/f=0$. It follows that $f$ has a holomorphic logarithm.