Is it true that every holomorphic vector bundle over $\mathbb{C}^{n}\setminus 0$ is trivial? If not true, how can one construct a counterexample? 


And just a small note here (**wrong**):


* *For $n\leq 2$, we can push the bundle $\mathcal{F}$ over $\mathbb{C}^{n}\setminus 0$ to be defined on $\mathbb{C}^{n}$, denoted by $\bar{\mathcal{F}}$. By taking a double dual $\bar{\mathcal{F}}^{\*\*}$ from which we can get reflexive sheaf which is isomorphic to $\mathcal{F}$ away from $0$. For a coherent analytic reflexive sheaf, we know that the set where it fails to be locally free has codimension $\geq 3$. So for $n\leq 2$, $\bar{\mathcal{F}}^{\*\*}$ is indeed a bundle. But for $n\geq3$, I have no clue.*


**This note turns to be wrong because I mistook $\bar{\mathcal{F}}$ to be a coherent sheaf, while the behavior of $\mathcal{F}$ near 0 is unknown.** 


**A further comment here:** For the line bundle case, based on the reference provided in the comments by pgraf, for $q\geq 1$, $H^{q}(\mathbb{C}^{n}\setminus 0, \mathcal{O})\neq 0$ iff $q=n-1$ which implies there exists nontrivial line bundle on $\mathbb{C}^{2}\setminus 0$ which cannot be extended to be on $\mathbb{C}^{2}$. In other cases, all holomorphic line bundles over $\mathbb{C}^{n}$ for $n\geq 3$ and $n=1$ would be trivial. Note when $n=1$, $\mathbb{C}^{\*}$ is a Stein manifold, or we can directly solve $\bar{\partial}$ equation by integration. 



Holomorphic vector bundles over $\mathbb{C}^{n}\setminus 0$
For $n\geq 3$, there exists such non-trivial bundles. For example, take the bundle on $\mathbb{C}^n-\{0\}$ given by the quotient of $\mathcal{O}^n$ by the subbundle generated by the vector $(x\_1,\ldots,x\_n)$, where the $x\_i$s are the coordinate functions. Then, for $n\geq 3$, this bundle is not trivial. The non-triviality will essentially follow from an argument like Hartog's theorem.


