I am unsure about a step in this proof. We define an analytic function to be $f:G\to C$ holomorphic and 1-1. Show that that its derivative is never zero.
Assume by contradiction that its derivative zero at some $a \in G$. Fiddling with th Taylor series expansion of $f$ gives $$h(z):=f(z)-f(a) = (z-a)^mg(z) $$ with $m >= 2$, $g(z)$ holomorphic and non zero at $a$.
Then $$e^{1/m\log(g(z))}=(g(z))^{1/m}$$ which gives that $f(z)=h(z)+f(a)=f(a)+((z-a)(g(z))^{1/m})^m$.
$((z-a)(g(z))^{1/m})^m$ is not 1-1 contradicting that $f$ is 1-1. But I do not see what makes $((z-a)(g(z))^{1/m})^m$ not 1-1.
By the Open Mapping Theorem, the image of the holomorphic function $(z-a) g(z)^{1/m}$ contains a neighbourhood of $0$. Thus for $|w|$ small enough, it contains all $m$ of $e^{2\pi i j/m} w$, $j = 0 \ldots m-1$. But the $m$'th power maps all these to the same point $w^m$, so this is not $1-1$.
If you don't know the Open Mapping Theorem, you can get it from the Argument Principle. Take $r$ small enough that the closed disk $D_r$ of radius $r$ around $a$ is contained in $G$ and such that $ g(z)$ is never $0$ there. Let $\Gamma$ be the positively oriented boundary of this disk. The number of zeros of $F(z)-w = (z-a) g(z)^{1/m} - w$ in $D_r$, counted by multiplicity, is given by an integral $$ \frac{1}{2\pi i} \oint_\Gamma \frac{F(z)-w}{F'(z)}\; dz$$ that is holomorphic for $w$ sufficiently close to $0$, and therefore is $1$ for such $w$.