We have given some complex analytic function $f:\mathbb{C} \to \mathbb{C}$.
I have read in a proof that if $f$ has exactly one fixed point $v \in \mathbb{C}$, then follows that $f$ is a polynomial and $f(z) = c(z-v)^n$ for some $c \in \mathbb{C}$ and $n \in \mathbb{N}$.
If I restrict $f$ to an arbitrary polynomial, this statement is clear to me. But why there are no transcendental functions which have exactly one fixed point?
This is not true. Take $f(z)=z(e^z+1)$. Obviously, $0$ is a fixed point. But, if $z\neq0$, then\begin{align}f(z)=z&\iff z(e^z+1)=z\\&\iff e^z+1=1\\&\iff e^z=0,\end{align}which is not possible. So, $0$ is the only fixed point.