Strictly upper triangular matrices form a nilpotent Lie algebra

For $p\ge 1$, let $F_p$ be the span of $E_{ij}$ with $j-i\ge p$; thus $F_1={\mathfrak{n}}_n$ and $F_n=0$. The identity $E_{ij}{E}_{\mathrm{\ell }m}={\delta }_{j\mathrm{\ell }}{E}_{im}$ shows that $[F_p,F_q]\subseteq F_{p+q}$: either product is zero, or its surviving matrix unit has superdiagonal distance equal to the sum of the two input distances.

Induction using [L1] and step 1.1 gives ${\gamma }_r({\mathfrak{n}}_n)\subseteq F_r$. Conversely, $F_1={\gamma }_1$, and if $r\ge 2$ and $E_{ij}\in F_r$, then $j-i\ge r$ and $E_{ij}=[E_{i,i+1},E_{i+1,j}]$; induction puts $E_{i+1,j}\in F_{r-1}={\gamma }_{r-1}$, so $E_{ij}\in {\gamma }_r$. Hence ${\gamma }_r({\mathfrak{n}}_n)=F_r$ for every $r\ge 1$.

In particular, ${\gamma }_n=F_n=0$, while ${\gamma }_{n-1}=F_{n-1}=k{E}_{1n}\mathrm{\ne }0$. Equivalently, the explicit left-normed chain $[E_{12},E_{23},\dots ,E_{n-1,n}]=E_{1n}$ witnesses sharpness. Thus [L2] gives class $n-1$. When $n=2$, the chain has one entry and ${\mathfrak{n}}_2=k{E}_{12}$ is abelian of class one; for $n=1$, outside the stated range, ${\mathfrak{n}}_1=0$ has class zero. No choice is used.