nth roots and logarithm Suppose $\Omega$ is a region,$f\in H(\Omega)$, and $f\not\equiv 0$.If for every positive integer n,there is $f_n \in H(\Omega)$,$(f_n)^n=f$,prove that there is $g\in H(\Omega),f=e^g$.
My conjecture :
$$g=lim  (nf_n - n)$$
But I don't know how to deal with it.
I will appreciate your help.
 A: If $f$ has a holomorphic logarithm $g$, then $f$ has no zeros, and $f' = g'\cdot e^g$, or
$$g' = \frac{f'}{f},$$
so the function $f'/f$ has a primitive on $\Omega$. Conversely, if $f$ has no zeros, and $f'/f$ has a primitive $h$ on $\Omega$, then $f\cdot e^{-h}$ is constant, and by adding a suitable constant to $h$, we obtain a $g$ with $f = e^g$.
So first, we deduce that $f$ has no zeros in $\Omega$: If $f$ had a zero in $a \in \Omega$, say of order $k$, then the $n$-th root $f_n$ would also have have a zero in $a$. Let its order be $k_n$, so $f_n(z) = (z-a)^{k_n}\cdot h_n(z)$ with $h_n \in H(\Omega)$ such that $h_n(a) \neq 0$. But then $f(z) = (z-a)^{n\cdot k_n}h_n(z)^n$, so $k = n\cdot k_n$, and $n \mid k$. But a positive integer has only finitely many divisors, so $f$ can't have a zero.
Thus $q = f'/f \in H(\Omega)$, and $q$ has a primitive if and only if
$$I(\gamma) := \int_\gamma \frac{f'(z)}{f(z)}\,dz = 0$$
for all closed paths $\gamma$ in $\Omega$. But $n(\gamma) = \frac{1}{2\pi i}I(\gamma)$ is the winding number of the closed path $f\circ \gamma$ around $0$, hence an integer, and
$$\begin{align}
n(\gamma) &= \frac{1}{2\pi i}\int_\gamma \frac{f'(z)}{f(z)}\,dz\\
&= \frac{1}{2\pi i} \int_\gamma \frac{(f_n(z)^n)'}{f_n(z)^n}\,dz\\
&= \frac{1}{2\pi i}\int_\gamma \frac{n f_n(z)^{n-1}f_n'(z)}{f_n(z)^n}\,dz\\
&= \frac{n}{2\pi i}\int_\gamma \frac{f_n'(z)}{f_n(z)}\,dz
\end{align}$$
is $n$ times the winding number of the closed path $f_n\circ\gamma$ around $0$, hence a multiple of $n$. The only integer that is a multiply of all $n \in \mathbb{Z}^+$ is $0$, hence $n(\gamma) = 0$, and indeed $f'/f$ has a primitive on $\Omega$.
