Proof that 1-1 analytic functions have nonzero derivative
I like proving this theorem via its contrapositive rather than by contradiction (though the computations are essentially the same).
Suppose $f:\Omega\to\mathbb{C}$ is analytic with $f'(z_0)=0$. The goal is to show that every disc about the origin contains distinct $z_1,z_2$ with $f(z_1)=f(z_2)$. We may assume that $z_0=f(z_0)=f'(z_0)=0$ (using $f(z+z_0)-f(z_0)$ if necessary, as translation doesn't affect injectivity). Since $f$ is analytic at $z=0$ and $f'(0)=f(0)=0$, $f$ has a power series expansion $$f(z)=a_k z^k + a_{k+1} z^{k+1} + \dots$$ where $k>1$. Pulling out a $z^k$ gives $$f(z)=z^k (a_k + a_{k+1}z + \dots) = z^k g(z)$$ where $g$ is analytic with $g(0)\neq0$. Since $g$ is nonzero on a sufficiently small disc centered at the origin, we can define an appropriate branch its log so that its kth root is well-defined. Call this function $h$, so that $h$ is analytic with $h(z)^k=g(z)$ near the origin. Hence $$f(z)=\left(zh(z)\right)^k.$$ Note that $\phi(z)=zh(z)$ is analytic (near $z=0$). Therefore, for any $\epsilon>0$ (sufficiently small), $\phi(D(0,\epsilon))$ is open (by the Open Mapping Theorem) and hence contains a disc $D(0,2\delta)$. In particular, there exist $z_1,z_2\in D(0,\epsilon)$ with $\phi(z_1)=\delta$ and $\phi(z_2)=\delta \exp\left(\frac{2\pi i}{k}\right)$. Therefore $$f(z_2)=\delta^k \exp\left(\frac{2\pi i}{k}\right)^k = \delta^k = f(z_1)$$ as desired.
The proof does look a bit clunky, especially with all the auxilliary functions. However, it's actually fairly simple and the extra functions are really just to show why each step is valid. In fact, the gist of the proof is:
Show that $f$ is the kth power of some analytic function $\phi$
Show that you can always find $z_1,z_2$ where $\phi(z_1)$ and $\phi(z_2)$ lie on the same circle and their arguments differ by $\frac{2\pi}{k}$, so that their k-th powers are equal.
Suppose $f$ is analytic at $z_0$ and non-constant, but $f'(z_0) = 0$. Then the order of the zero of $f(z) - f(z_0)$ at $z_0$ is some integer $k > 1$. Take some circle $\Gamma$ around $z_0$ so $f$ is analytic on and inside $\Gamma$ and there are no other zeros of $f(z) - f(z_0)$ or $f'(z)$ on or inside $\Gamma$. Now the sum of the orders of the zeros of $f(z) - p$ inside $\Gamma$ is $\dfrac{1}{2\pi i}\int_\Gamma \frac{f'(z)}{f(z) - p}\, dz.$, which is equal to $k$ for $p$ in a neighbourhood of $f(z_0)$. But since $f'(z)$ has no other zeros inside $\Gamma$, those zeros are simple, i.e. there are $k$ distinct solutions to $f(z) = p$ inside $\Gamma$.