$f: \Omega \rightarrow \Omega$ holomorphic, $f(0) = 0$, $f'(0) = 1$ implies $f(z) = z$

Solution 1:

A solution which I find more natural than looking at iterations of $f$ involves a holomorphic covering map $\phi:\mathbb D\to\Omega$. Such a map exists for every domain $\Omega$ provided that $\mathbb C\setminus \Omega$ has at least two points. This is Theorem 16.5.1 in Conway's book Functions of one complex variable, vol. II.

Normalize the covering map so that $\phi(0)=0$. Then $\phi^{-1}\circ f\circ \phi$ is a holomorphic map of disk to itself which fixes the origin and has derivative $1$ there. (Note: although $\phi$ is not globally invertible, it is locally invertible; therefore $\phi^{-1}\circ f\circ \phi$ is an analytic function, and since $\mathbb D$ is simply-connected, it has a single-valued holomorphic branch in $\mathbb D$.)

By the Schwarz lemma the map $\phi^{-1}\circ f\circ \phi$ is the identity, and therefore $f$ is the identity.

Key words for additional reading: uniformization theorem, hyperbolic metric.