 A univalent function has nonzero derivative

For an injective holomorphic function $f$ on an open subset of $\mathbb{C}$, $f^{\prime }(z)$ is nowhere zero. If $f(z)-f(z_0)=(z-z_0{)}^{m}h(z)$ with $h(z_0)\ne 0$, choose a small circle where both $h$ and $mh+(z-z_0)h^{\prime }$ have no zeros. The argument principle counts $m$ zeros of $f(z)-f(z_0)-w$ inside that circle for sufficiently small $w\ne 0$: deforming $w$ to zero cannot change the count. All these zeros are simple, since $f^{\prime }$ only vanishes at $z_0$ inside the circle and $z_0$ is not a zero when $w\ne 0$. Injectivity forces $m=1$, and hence $f^{\prime }(z_0)\ne 0$.