Let $L$ be a Lie algebra. Is $Z(L) \subset [L,L]$?

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In a proof showing that if $L$ is nilpotent then $L/Z(L)$ is nilpotent, the author begins with: "Say $L^n \subset Z(L)$".

I'm assuming he's writing this since we're assuming $(L/Z(L))^n = 0$ for some $n$.

I believe he is implying $(L/Z(L))^n=L^n/Z(L)$?

I'm trying to show at least that $L^1/Z(L)$ is well-defined. Thus, trying to show that $L^1=[L,L]$ contains $Z(L)$.

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For a nilpotent Lie algebra, the descending series defined by $L^1=L$ and $L^i=[L,L^{i-1}]$ ends with $0$, i.e., we have $L^{n}=0$ for some $n$. The last term before, which is $L^{n-1}$, then clearly is contained the center because of $[L^{n-1},L]=L^n=0$. However, it is not true in general that we have $n=2$, i.e., $Z(L)\subseteq [L,L]$. Counterexamples are abelian Lie algebras, or, for example, $L=\mathfrak{n}_3\oplus K$ with basis $(x_1,x_2,x_3,x_4)$ and brackets $[x_1,x_2]=x_3$. Then $Z(L)=\langle x_3,x_4\rangle$, but $[L,L]=\langle x_3\rangle$.

In particular, for a nilpotent Lie algebra $L$ we can say "Say $L^n\subseteq Z(L)$".