= A univalent function has nonzero derivative
For an <injective> <holomorphic function> $f$ on an <open subset> of $\mathbb C$, $f'(z)$ is nowhere zero. If $f(z)-f(z_0)=(z-z_0)^m h(z)$ with $h(z_0)\ne0$, choose a small <circle> where both $h$ and $mh+(z-z_0)h'$ have no zeros. The <argument principle> counts $m$ zeros of $f(z)-f(z_0)-w$ inside that <circle> for sufficiently small $w\ne0$: deforming $w$ to zero cannot change the count. All these zeros are simple, since $f'$ only vanishes at $z_0$ inside the <circle> and $z_0$ is not a zero when $w\ne0$. Injectivity forces $m=1$, and hence $f'(z_0)\ne0$.
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