I recently saw a lecturer prove the following theorem (assuming the result that every analytic function is locally 1-1 whenever its derivative is nonzero): Let $\Omega \subset \mathbb{C}$ be open, and let $f : \Omega \to \mathbb{C}$ be 1-1 and analytic on $\Omega$. Then $f'(z_0) \not = 0$ for every $z_0 \in \Omega$.
I got the basic idea behind the proof: we assume for contradiction that $f'(z_0) = 0$, and, assuming without loss of generality that $z_0 = f(z_0) =0$, we have (from the power-series expansion) that $f(z) = z^kg(z)$ for some analytic $g$ in some disk at the origin (i.e., $z_0$) and some $k \ge 2$. Since $z^k$ is not 1-1 in any such disk (because there are multiple roots of unity), then $f$ isn't either.
However, the proof he gave was rather awkward and technical- it involved defining three different axillary functions, even though the idea was simple, and I've since forgotten how it exactly worked. In any case, I'm convinced there's a better way.
The problem is that I'm having trouble turning the idea into a real proof- I know that it obviously follows if $g$ is 1-1, but I'm also pretty sure that that is too strong an assumption. Am I missing something, or does the argument just have to be more complicated?