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.