Why is every holomorphic line bundle over $\mathbb{C}$ trivial? Why is every holomorphic line bundle over $\mathbb{C}$ necessarily trivial? 
I am having a hard time finding a proof to this seemingly innocuous fact. 
I have tried showing that such a line bundle will have a nowhere-zero global section (this is one of the few criterion I currently have at my disposal for showing a line bundle is trivial). 
I would much appreciate either a link to a proof of this statement or a hint of how I could prove this myself. 
 A: Let $X$ be a complex manifold and denote by $\operatorname{Pic}(X)$ the Picard group of $X$ which is the collection of isomorphism classes of holomorphic line bundles on $X$ with binary operation given by tensor product and the isomorphism class of the trivial bundle serves as the identity element. By considering transition functions, one can show that $\operatorname{Pic}(X) \cong H^1(X, \mathcal{O}^*)$ (as groups).
The exponential sheaf sequence 
$$0 \to \mathbb{Z} \to \mathcal{O} \to \mathcal{O}^* \to 0$$
gives rise to a long exact sequence in cohomology
$$\dots \to H^1(X, \mathbb{Z}) \to H^1(X, \mathcal{O}) \to H^1(X, \mathcal{O}^*) \to H^2(X, \mathbb{Z}) \to \dots$$
Now consider the case $X = \mathbb{C}$. As $\mathbb{C}$ is contractible, $H^1(\mathbb{C}, \mathbb{Z}) = 0$ and $H^2(\mathbb{C}, \mathbb{Z}) = 0$. Therefore, 
$$\operatorname{Pic}(\mathbb{C}) \cong H^1(\mathbb{C}, \mathcal{O}^*) \cong H^1(\mathbb{C}, \mathcal{O}) \cong H^{0,1}_{\bar{\partial}}(\mathbb{C})$$
where the last isomorphism is given by Dolbeault's Theorem. 
The final piece of the puzzle is the following non-trivial fact: for any smooth function $f$ on $\mathbb{C}$, there is a smooth function $u$ such that $\frac{\partial u}{\partial \bar{z}} = f$. As any $(0,1)$-form on $\mathbb{C}$ is of the form $f d\bar{z}$, we see that all $(0, 1)$-forms are $\bar{\partial}$-exact: $f d\bar{z} = \frac{\partial u}{\partial \bar{z}} d\bar{z} = \bar{\partial}u$. Therefore, $H^{0,1}_{\bar{\partial}}(\mathbb{C}) = 0$ and hence every holomorphic line bundle on $\mathbb{C}$ is trivial.
