Given linear transformations $S,T:V→V$, prove that if $x=0$ is an eigenvalue of $ST$, then it is an eigenvalue of $TS$.
Secondly, prove that if $y\neq 0$ is an eigenvalue of $ST$, and that if $v$ is eigenvector of $ST$ for $y$, then $Tv$ is an eigenvector of $TS$ for $y$.
For this first part, there may be a simpler approach, but this is what came to mind. Since $ST$ has zero as an eigenvalue, we have $$\det S \det T = \det ST = 0.$$ So, either $S$ or $T$ (or both) is singular. If $S$ is singular, take $v_0 \neq 0$ so that $Sv_0=0$. Then, $$TSv_0 = T\vec{0} = \vec{0}.$$ Suppose $T$ is singular. If $S$ is also singular, then, by the above, we are done. Suppose $S$ is not singular. Take $v_0 \neq 0$ so that $Tv_0 = 0$ and pick $w_0$ so that $Sw_0=v_0$.
For the next part, $$TS(Tv) = T(STv) = T(yv) = yTv,$$ which implies that $Tv$ is an eigenvector of $TS$ with corresponding eigenvalue $y$.