Dear David, I think there might be a slightly simpler proof of the analytic Quillen-Suslin theorem.

Given a holomorphic vector bundle $E$ on $\mathbb C^n$, it has a holomorphic connection since its Atiyah class $a(E)$ vanishes : indeed $a(E) \in H^1 (\mathbb C^n, \Omega^1_X\otimes \mathcal {End}E) $ , and the whole cohomology group is zero by Theorem B. And then I remember that a long tile ago, Otto Forster explained in a talk an argument by Griffiths that you could deduce from that connection the holomorphic triviality of $E$. Extend each vector $v\in E[0]$, the fiber of $E$ at zero , to to a global section $s_v \in \Gamma (\mathbb C^n, E)$ by first doing that on the restriction $E|L$ of $E$ to lines $L\subset \mathbb C^n$ and then show holomorphic dependency on $v$ by invoking a theorem of holomorphic dependency of a system of differential equations on its paramaters (here $v$).This gives a trivialization of $E$. Unfortunately I don't have a reference for this method of proof but I have no doubt you can fill in the details if you feel so inclined. However it is not clear how elementary this proof really is, since it invokes theorem B .

Finally, since you evoke the contrast for a manifold between having its holomorphic line bundles trivial and all its holomorphic vector bundles of any rank trivial, let me quote :
Theorem (Forster-Rammspott) On a Stein manifold of dimension $n$ every holomorphic vector bundle $E$ of rank $r\geq [n/2]$ can be decomposed as $E=F\oplus \mathcal O^{r-[n/2]}$ where $F$ is a holomorphic vector bundle of rank $ [n/2]$ .

So we have the surprizing consequence that for a Stein surface the hypothesis that all holomorphic line bundles are trivial forces the conclusion that all holomorphic bundles, of any rank, are trivial.

Dear David, I think there might be a slightly simpler proof of the analytic Quillen-Suslin theorem.

Given a holomorphic vector bundle $E$ on $\mathbb C^n$, it has a holomorphic connection since its Atiyah class $a(E)$ vanishes : indeed $a(E) \in H^1 (\mathbb C^n, \Omega^1_X\otimes \mathcal {End}E) $ , and the whole cohomology group is zero by Theorem B. And then I remember that a long tile ago, Otto Forster explained in a talk an argument by Griffiths that you could deduce from that connection the holomorphic triviality of $E$. Extend each vector $v\in E[0]$, the fiber of $E$ at zero , to to a global section $s_v \in \Gamma (\mathbb C^n, E)$ by first doing that on the restriction $E|L$ of $E$ to lines $L\subset \mathbb C^n$ and then show holomorphic dependency on $v$ by invoking a theorem of holomorphic dependency of a system of differential equations on its paramaters (here $v$).This gives a trivialization of $E$. Unfortunately I don't have a reference for this method of proof but I have no doubt you can fill in the details if you feel so inclined. However it is not clear how elementary this proof really is, since it invokes theorem B .

Finally, since you evoke the contrast for a manifold between having its holomorphic line bundles trivial and all its holomorphic vector bundles of any rank trivial, let me quote :
Theorem (Forster-Rammspott) On a Stein manifold of dimension $n$ every holomorphic vector bundle $E$ of rank $r\geq [n/2]$ can be decomposed as $E=F\oplus \mathcal O^{r-[n/2]}$ where $F$ is a holomorphic vector bundle of rank $ [n/2]$ .

So we have the surprizing consequence that for a Stein surface the hypothesis that all holomorphic line bundles are trivial forces the conclusion that all holomorphic bundles, of any rank, are trivial.

An example (edit) In his comment Johannes remarks that on a non-compact Riemann surface, all holomorphic vector bundles are trivial. This is no longer true in higher dimensions in general : $\mathbb C^n $ is special in that respect.
For example take $X=\mathbb P^2(\mathbb C) \setminus C$ where $C$ is the conic $x^2+y^2+z^2=o$ . The surface $X$ is Stein (even affine) but its tangent bundle is not holomorphically trivial because it is not even topologically trivial. Since holomorphic line bundles are classified by $H^2(X,\mathbb Z)=\mathbb Z /2$ , as David pointed out, the holomorphic line bundles on $X$ aren't trivial either ( which you could predict from Forster-Ramspott 's theorem !).

Dear David, I think there might be a slightly simpler proof of the analytic Quillen-Suslin theorem.

Given a holomorphic vector bundle $E$ on $\mathbb C^n$, it has a holomorphic connection since its Atiyah class $a(E)$ vanishes : indeed $a(E) \in H^1 (\mathbb C^n, \Omega^1_X\otimes \mathcal {End}E) $ , and the whole cohomology group is zero by Theorem B. And then I remember that a long tile ago, Otto Forster explained in a talk an argument by Griffiths that you could deduce from that connection the holomorphic triviality of $E$. Extend each vector $v\in E[0]$, the fiber of $E$ at zero , to to a global section $s_v \in \Gamma (\mathbb C^n, E)$ by first doing that on the restriction $E|L$ of $E$ to lines $L\subset \mathbb C^n$ and then show holomorphic dependency on $v$ by invoking a theorem of holomorphic dependency of a system of differential equations on its paramaters (here $v$).This gives a trivialization of $E$. Unfortunately I don't have a reference for this method of proof but I have no doubt you can fill in the details if you feel so inclined. However it is not clear how elementary this proof really is, since it invokes theorem B .

Finally, since you evoke the contrast for a manifold between having its holomorphic line bundles trivial and all its holomorphic vector bundles of any rank trivial, let me quote :
Theorem (Forster-Rammspott) On a Stein manifold of dimension $n$ every holomorphic vector bundle $E$ of rank $r\geq [n/2]$ can be decomposed as $E=F\oplus \mathcal O^{r-[n/2]}$ where $F$ is a holomorphic vector bundle of rank $r\geq [n/2]$ .

So we have the surprizing consequence that for a Stein surface the hypothesis that all holomorphic line bundles are trivial forces the conclusion that all holomorphic bundles, of any rank, are trivial.

Dear David, I think there might be a slightly simpler proof of the analytic Quillen-Suslin theorem.

Given a holomorphic vector bundle $E$ on $\mathbb C^n$, it has a holomorphic connection since its Atiyah class $a(E)$ vanishes : indeed $a(E) \in H^1 (\mathbb C^n, \Omega^1_X\otimes \mathcal {End}E) $ , and the whole cohomology group is zero by Theorem B. And then I remember that a long tile ago, Otto Forster explained in a talk an argument by Griffiths that you could deduce from that connection the holomorphic triviality of $E$. Extend each vector $v\in E[0]$, the fiber of $E$ at zero , to to a global section $s_v \in \Gamma (\mathbb C^n, E)$ by first doing that on the restriction $E|L$ of $E$ to lines $L\subset \mathbb C^n$ and then show holomorphic dependency on $v$ by invoking a theorem of holomorphic dependency of a system of differential equations on its paramaters (here $v$).This gives a trivialization of $E$. Unfortunately I don't have a reference for this method of proof but I have no doubt you can fill in the details if you feel so inclined. However it is not clear how elementary this proof really is, since it invokes theorem B .

Finally, since you evoke the contrast for a manifold between having its holomorphic line bundles trivial and all its holomorphic vector bundles of any rank trivial, let me quote :
Theorem (Forster-Rammspott) On a Stein manifold of dimension $n$ every holomorphic vector bundle $E$ of rank $r\geq [n/2]$ can be decomposed as $E=F\oplus \mathcal O^{r-[n/2]}$ where $F$ is a holomorphic vector bundle of rank $ [n/2]$ .

So we have the surprizing consequence that for a Stein surface the hypothesis that all holomorphic line bundles are trivial forces the conclusion that all holomorphic bundles, of any rank, are trivial.