The inverse of a bijective holomorphic function is also holomorphic

Step 1: $f'$ is never zero.

Indeed, if $f'(a)=0$ for some $a\in U$, then the Taylor expansion at $a$ is of the form $f(z)=f(a) + c_n (z-a)^n+\dots$ with $n\ge 2$, $c_n\ne 0$. This implies that $g(z) = (f(z)-f(a))/(z-a)^n$ is a nonzero holomorphic function near $a$, hence admits an $n$th degree root (a function $h$ such that $h^n=g$). Hence $$f(z) = f(a) + [(z-a)h(z)]^n$$ Since $z\mapsto (z-a) h(z) $ is an open map, its image contains a neighborhood of $0$; in particular it contains the points $\epsilon$ and $\epsilon \exp(2\pi i /n)$ for small $\epsilon$. These two points are sent into one, contradicting the injectivity of $f$.

Step 2: Inverse is smooth

This is just the Inverse function theorem: writing out $f=u+iv$, one can see that the Jacobian determinant of $(x,y)\mapsto (u,v)$ is $|f'(z)|^2\ne 0$.

Step 3: Inverse is holomorphic

Also the Inverse function theorem. Writing the derivative of $f$ as a $2\times 2$ real matrix, we get something of the form $$\begin{pmatrix} a & b \\ -b& a\end{pmatrix}$$ due to the Cauchy-Riemann equations. The inverse of such a matrix is also of this form: hence, $f^{-1}$ satisfies the Cauchy-Riemann equations.

Hint: Prove that $f^{\prime}(z)\neq 0$ for all $z\in U$. Then use the inverse function theorem for analytic functions.