Exists there a biholomorphic function $f\colon\mathbb{C}\to\mathbb{C}\setminus \{0\}$?
My idea:
Suppose, there exists a biholomorphic function $f\colon\mathbb{C}\to\mathbb{C}\setminus \{0\}$. Because of $f(z)\neq 0 \ \forall z\in\mathbb{C}$ there exists $\delta>0$ with $|f(z)|\geq \delta\ \forall z \in\mathbb{C}$.
Define $g\colon\mathbb{C} \to \mathbb{C}, g(z)=\frac{1}{f(z)}.$ Then $g$ is holomorphic and bounded by $\frac{1}{\delta}$. Because of Liouville $g$ (and so $f$) is constant: contradiction!
Is this correct?
Edit:
$\mathbb{C}$ is simply-connected.
Suppose, $f$ is biholomorphic.
Then $f(\mathbb{C})=\mathbb{C}\setminus\{0\}$ is simply-connected.
This is a contradiction because $\mathbb{C}\setminus\{0\}$ is not simply-connected.
Best Answer
Let $\gamma$ be the image of the unit sphere $\partial B_1(0)$ under the proposed inverse $g: \mathbb C^\times \longrightarrow \mathbb C$ of your biholomorphic map $f : \mathbb C\longrightarrow \mathbb C^\times$. Then prove and study the consequences of the equality of integrals
$$ \frac 1 {2\pi i}\int_\gamma \frac{f'(w)}{f(w)} dw =\frac 1 {2\pi i}\int_{\partial B_1(0)}\frac {dz} z $$
by observing that $f'/f$ is holomorphic on $\mathbb C$.