Let $f:U \to V$ be a bijective holomorphic function. Show that inverse of $f$ is also holomorphic. Suppose $U$ and $V$ be domains(i.e., open and connected) in $ \mathbb C$.Let $f\colon U \to V$ be a bijective holomorphic function. Show that the inverse of $f$ is also holomorphic.
By Open Mapping Theorem it is clear that $f^{-1}$ is also continuous. Please give some ideas to complete the proof.
Edit:I'm interested in a proof which comes as a corollary of  open mapping theorem
 A: By the open mapping theorem, $f$ is a homeomorphism. When $f'(z_0) \neq 0$, the complex differentiability of $f^{-1}$ at $f(z_0)$ follows in the usual way.
Now consider $U' = \{ z\in U : f'(z) \neq 0\}$ and $V' = f(U')$. By the above, $f^{-1}$ is holomorphic on $V'$. But since $U\setminus U'$ consists only of isolated points, and $f$ is a homeomorphism, $V\setminus V'$ consists only of isolated points. Thus a $w\in V\setminus V'$ would be an isolated singularity of $f^{-1}$. Since $f^{-1}$ is continuous, it would be a removable singularity. Hence $f^{-1}$ is holomorphic on all of $V$. (And consequently we have $U' = U$.)
A: Hint: What happens in a small disk around $z_0\in U$ if $f'(z_0)=0$?
A: Following the hint by Hagen von Eitzen, we can show that, if there is a $z_0$ such that $f'(z_0)=0$, then $f$ is not injective. Consider a small circular path $\gamma$ around $z_0$. By Rouche's theorem, the winding number of $f(\gamma)$ around $f(z_0)$ is the order of the zero $z_0$ of $f(z)-f(z_0)$ which, by inspecting its Taylor series at $z=z_0$, is greater than 1. So $f(\gamma)$ must cross itself somewhere, making $f$ not injective. By contraposition, $f'$ is never 0. Now we have that $f^{-1}(z)$ is holomorphic because it's differentiable and \begin{eqnarray} \frac{df^{-1}(z)}{dz}=\frac{1}{f'(f^{-1}(z))}\end{eqnarray}
