f'(z)<>0

[Солунац]

Let $f: G\rightarrow \mathbb{C}$ is analytic and one-to-one, where $G$ is open subset of $\mathbb{C}$. Prove that $\forall z\in G$ is $f'(z)\neq 0$.

Posted after 4 minutes 18 seconds:

I have a hint that if $n = ord(g;z_0) > 0$ for some $z_0\in G$, then there is an open neighbourhood $z_0\in U\subset G$ and an analytic function $h: U\to \mathbb C$ such that $g(z) = h(z)^n$ for all $z\in U$.

But, problem is that I don't know why then exists such analytic function, and, after all what to do with it. Sorry...

Спасибо.

[Полосин]

Suppose $f'(z_0)=0$ and consider the Taylor series at $z_0$.