$L_1,L_2$ linear and $L_1\circ L_2$ nilpotent $\implies L_2\circ L_1$ nilpotent

Question

Given $L_1,L_2\in\mathcal{L}(V)$ and $L_1L_2$ nilpotent. I have to show that $L_2L_1$ ist nilpotent either.
Let $L_1,L_2\in\mathcal{L}(V)$ and $L_1L_2$ nilpotent. Let $v\in V$. Than there $\exists m\in\mathbb{N}: (L_1L_2)^m(v)=0$

Is it true that $(L_1L_2)^m(v)=0 \Leftrightarrow L_1^m(L_2(v))=0$?
If so than $(L_1L_2)$ nilpotent would imply $L_1$ nilpotent and I can go on by myself. If not... I could use some help.

Thanks, Matthias

Answer 1

If $(L_1L_2)^n = 0$, then $(L_2L_1)^{n+1} = L_2(L_1L_2)^nL_1 = 0$