For some open sets $U$, $V$ in the complex plane, let $f:U\rightarrow V$ be an injective holomorphic function. Then $f'(z) \ne 0$ for $z \in U$.
Now I don't understand the proof, but here it is from my text. My comments are in italics.
Suppose $f(z_0) = 0$ for some $z_0 \in U$.
$f(z) - f(z_0) = a(z - z_0)^k + G(z)$ for all $z$ near $z_0$, with $a \ne 0, k \ge 2.$ Also, $G$ vanishes to order $k+1$ at $z_0$.
I'm not clear on what this "vanishing" thing means. Maybe it means that $G$ can be expressed as a power series of order $k+1$ around $z_0$.
For sufficiently small $w$ we can write $f(z) - f(z_0) - w = F(z) + G(z)$, where $F(z) = a(z - z_0)^k - w$.
I'm not sure why we need to have $w$ small. This equation will work for any $w$.
Since $|G(z)| \lt |F(z)|$ on a small circle centered at $z_0$, and $F$ has at least two zeros inside that circle, Rouche's theorem implies that $f(z) - f(z_0) - w$ has a least two zeros there.
Now I think that $|G(z)| \lt |F(z)|$ can follow simply from the fact that $F$ is a polynomial of degree $k$ while $G$ has degree $k+1$. And the remark about the two zeros can follow from the fact that $F$ must have $k$ zeros in the complex plane. But the first part requires that we consider $z$ only on a small circle. The second part requires that our circle be big enough to capture two zeros. How do we know that we can satisfy both?
Since $f'(z) \ne 0$ for for all $z \ne z_0$ sufficiently close to $z_0$, the roots of $f(z) - f(z_0) - w$ are distinct, so $f$ is not injective - a contradiction.
I think that the derivative is never zero for values of $z$ other than $z_0$ because otherwise we would have a sequence of zeros limiting towards $z_0$ which would cause our function to be constant which is a contradiction. But again we have the same problem - we can only consider a small circle. The roots of $f$ may lie outside this circle.