If $\,f^{7} $ is holomorphic, then $f$ is also holomorphic. I need some help with this problem:


  Let $ \Omega $ be a complex domain, i.e., a connected and open non-empty subset of $ \mathbb{C} $. If $ f: \Omega \to \mathbb{C} $ is a continuous function and $ f^{7} $ is holomorphic on $ \Omega $, then $ f $ is also holomorphic on $ \Omega $.


Thanks in advance.
We may assume $\Omega$ connected.

Let $g(z)\colon = [f(z)]^7$, a holomorphic function. Let $N$ the set of zeroes of $g$. If $N$ has an  accumulation point  in $\Omega$ then $g$ is identical $0$ and so is $f$. 
Otherwise, the set $N$ is discrete ( it may have an accumulation point in $\partial \Omega$).

The set $\Omega \backslash N$ is open and connected. The function $g$ maps $\Omega \backslash N$  to $\mathbb{C}\backslash\{0\}$. Let $p \colon \mathbb{C}\backslash\{0\} \to \mathbb{C}\backslash\{0\}$, $p(z) = z^7$. 

$p$ is a holomorphic covering . We have 
$$p\circ f = g$$
that is, $f$ is a lift of $g$. Moreover, $f$ is continuous. It is a general fact that any continuous lift of a  holomorphic map for a holomorphic covering is also holomorphic. We conclude that $f$ is holomorphic on $\Omega \backslash N$. The singularities $N$ are removable because $f$ is continuous on $\Omega$. So $f$ holomorphic on $\Omega$.

Obs: In fact it works with any nonconstant holomorphic function $\phi(z)$ instead of $z^7$. The problem can be treated locally in a similar way.
There exist exactly $7$ holomorphic functions $g_0,...g_6$ such that $g_i^7=f^7$ (there are seven holomoprhic $7^{th}$ root functions which we can compose with $f^7$, or $7$ branches if you prefer). We need to show $f$ is equal to one of them. We know that everywhere it takes a value that is equal to one of the $7^{th}$ roots of $f^7$; Hence we can argue by continuity that its values lie on one of the branches only, and we are done.