Are strictly upper triangular matrices nilpotent?

An $n\times n$ matrix $A$ is called nilpotent if $A^m = 0$ for some $m\ge1$.

Show that every triangular matrix with zeros on the main diagonal is nilpotent.

Its characteristic polynomial is $T^n$, so by Cayley-Hamilton, $A^n=0$.

WLOG assume that $A$ is upper-triangular (otherwise, a similar argument works with the basis reversed).

Regard $A$ as a linear transformation on $\mathbb{F}^n$ with basis $e_1, \ldots, e_n$. Let $U_i$ be the span of $e_1, e_2, \ldots, e_i$ for $i = 0, 1, \ldots, n$. Observe that $0 = U_0 \subseteq \ldots \subseteq U_n = \mathbb{F}^n$ and also note that $AU_i \subseteq U_{i - 1}$ for $i = 1, \ldots, n$ since $A$ is strictly upper-triangular. Therefore, $A^n = 0$.