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.

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.
