The Wikipedia page on nilpotent matrices claims the following:
A nilpotent transformation $L$ on $\mathbb{R}^n$ naturally determines a flag of subspaces:
$${0} \subset \ker L \subset \ker L^2 \subset \dots \subset \ker L^q = \mathbb{R}^n$$
Intuitively this makes sense to me, but I haven't been able to find a formal proof.
I managed to show that $\ker L^{k-1} \subseteq \ker L^k$, but haven't been able to show that $\ker L^{k-1} \subset \ker L^k$ as required, and was wondering how to approach this.
Suppose $\ker L^{k-1}=\ker L^k$ for some $k$. I claim $\ker L^k=\ker L^{k+1}$.
Take $v\in\ker L^{k+1}$. Then $L^{k+1}(v)=0$, so $L^k(L(v))=0$, so $L(v)\in\ker L^k=\ker L^{k-1}$. Then $L^{k-1}(L(v))=L^k(v)=0$, so $v\in\ker L^k$. We get $\ker L^{k+1}\subseteq\ker L^k$, and equality follows.
Inductively, it follows $\ker L^{k-1}=\ker L^k=\ker L^{k+1}=\ker L^{k+2}=\cdots$. So if $\ker L^{k-1}$ is proper in $\mathbb{R}^n$, so is $\ker L^n$ for all $n$. But this is a contradiction since $L$ is nilpotent, so that $L^n=0$ for some $n$.