# Nilpotence of Lie Algebra

I am trying to show that if $L$ is a Lie algebra and $L/Z(L)$ is nilpotent than $L$ is also nilpotent. Can someone please help me?

I tried to first show by induction: $(L/Z(L))^k=L^k/Z(L)$. Is it correct?

-

EDIT: in order to actually answer your question: yes your formula is correct, if by $L^k/Z(L)$ you mean the image of $L^k$ under the canonical map $L\to L/Z(L)$.
$L/Z(L)$ nilpotent means that for some integer $N$, and all $x_1,\dots,x_N\in L,$ $$[x_1,[x_2,\dots[x_{N-1},x_N]\dots ]]\in Z(L).$$ Thus, for all $x_0, x_1,\dots,x_N\in L$, $$[x_0,[x_1,\dots[x_{N-1},x_N]\dots ]]=0,$$ i.e. $L^{N+1}=0$, and $L$ is nilpotent.