In the process of showing map $f: \mathbb{C} \to \mathbb{C}$ is biholomorphic. If we assume $f$ is injective and holomorphic, I want to use the little Picard theorem to illustrate $f$ surjective, but there is a stunk for me that if $$f=e^z: \mathbb{C} \longrightarrow \mathbb{C}^{*}=\mathbb{C}\setminus \{0\}$$
How to deal with it? Since Picard's little theorem supports the map is not just onto $\mathbb{C}$.
I think we're trying to show that if $f$ is an injective entire function then $f$ is surjective.
Suppose not; say $f(z)\ne\alpha$ for every $z$. Let $g=f-\alpha$. Then $g=e^h$ for some entire function $h$. Little Picard shows that $h$ must assume more than one of the values $0,2\pi i, 4\pi i,\dots$; hence $e^h$ is not injective.