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.

It's true. If the zeros of f have a limit point, then so is $f^7$, but $f^7$ is holomorphic, so $f^7$ is zero every where. If all zeros of f is isolated, we can apply the Cauchy-Riemann equation and use the chain rule to get $ f^6 \cdot \frac{\partial f}{\partial \bar{z}} = 0$. Thus $f$ is holomorphic at $z$ whenever $ z$ is not a zero of $f$. Since $f$ is continuous and its zeros are isolated, we know from the removable singularity theorem that $f$ is also holomorphic at these zeros.
