Proof of the linear independence of the generalized eigenvectors of a square matrix I'm currently stuck on this problem:

Let $V$ be a finite dimensional vector space. If $S: V\rightarrow V$ and $T: V\rightarrow V$ are linear maps and $ST=TS$, prove every eigenvalue of $ST$ is a product of an eigenvalue of $S$ with an eigenvalue of $T$.

One thing that I do know - as a consequence of the above - is that there is a basis of simultaneous eigenvectors. Do I somehow use this fact?
 A: You need the scalar field to be algebraically closed (see Berci's example in he comments above). I am pretty sure it is $\mathbb{C}$ in your case, but the argument works exactly the same for a general algebraically closed field $K$.
Fact: any $S\in L(V)$ is triangularizable. That is, there exists a basis of $V$ with respect to which the matrix of $T$ is upper triangular.

Generalization: any pair of commuting $S,T\in L(V)$ is simultaneously upper triangularizable.
Application: take a basis of simultaneous upper triangularization for $S$ and $T$. Then the matrix of $ST=TS$ is also upper triangular. Its diagonal is the product of the diagonals of $S$ and $T$. In particular, every eigenvalue of $ST$ is the product of an eigenvalue of $S$ by an eigenvalue of $T$.

Proof of the fact: induction on $n$, the dimension of $V$. The key point is that there always exists at least an eigenvalue $\lambda$, as the characteristic polynomial splits over $K$. For $n=1$, the assertion is trivial. Assume it holds up to $n-1\geq 1$. Then take an eigenvalue $\lambda$. Take a basis of the eigenspace $V_\lambda:=\mbox{Ker}(S-\lambda I)$, and complete it into a basis of $V$. If $V_\lambda$ was already equal to$ V$, this means $S$ is scalar and we are done. If not, then the matrix of $S$ with respect to this basis is a $2\times 2$ block matrix. The upper left block is $\lambda I$, and the lower left block is $0$. Apply the induction hypothesis to the lower right block. QED.
Proof of the generalization: you can do that by induction on $n\dim V$ again. On a one-dimensional space, it is trivial. Assume this holds for any dimension up to $n-1\geq $. Take $S,T$ commuting. Since $K$ is algebraically closed, the characteristic polynomial of $S$ has at least one root $\lambda$. This is an eigenvalue of $S$. On the eigenspace $V_\lambda:=\mbox{Ker}(S-\lambda I)$, $Sx=\lambda x$. Therefore $STx=TSx=\lambda Tx$, i.e. $Tx\in V_\lambda$ for every $x\in V_\lambda$. In other words, $V_\lambda$ is invariant under $T$. Using the fact, you get a basis of $V_\lambda$ in which $T$ is upper triangular and, of course, $S$ is diagonal. If $V_\lambda=V$, we are done. If not, take a basis of $V_\lambda$, and complete it into a basis of $V$. Then the matrices of $S$ and $T$ are $2\times 2$ block matrices. As $V_\lambda$ is invariant under $S$ and $T$, they are block upper triangular. We don't care about the upper right blocks. By writing down the corresponding $2\times 2$ block product, the condition $ST=TS$ implies that the lower right blocks are still commuting. Apply the induction hypothesis to them. QED.
A: There is only a basis of simultaneous eigenvectors if $S$ and $T$ are both diagonalizable, but in that case the answer is yes - try multiplying some element of the basis of simultaneous eigenvectors by $ST$ and seeing what happens. This will give you a set of eigenvalues of $ST$, and you need to see why this set must in fact be all the eigenvalues.
