Upper triangular matrices are solvable but not nilpotent

The diagonal entry of a commutator of two upper triangular matrices is zero, so ${\mathfrak{b}}_n^{(1)}$ lies in the strictly upper triangular subspace $F_1$. More generally, if $F_p$ is spanned by $E_{ij}$ with $j-i\ge p$, matrix-unit multiplication gives $[F_p,F_q]\subseteq F_{p+q}$ and $F_n=0$. Induction yields ${\mathfrak{b}}_n^{(r)}\subseteq F_{2^{r-1}}$ for $r\ge 1$, so a derived term vanishes once $2^{r-1}\ge n$. Thus ${\mathfrak{b}}_n$ is solvable by [L1].

Let $H=E_{11}$ and $X=E_{12}$. Then $[H,X]=X$, so $X\in {\gamma }_2({\mathfrak{b}}_n)$. If $X\in {\gamma }_r$, the same bracket puts $X$ in $[{\mathfrak{b}}_n,{\gamma }_r]={\gamma }_{r+1}$. Hence $X$ belongs to every ${\gamma }_r$ for $r\ge 2$, and none of these terms is zero. Therefore ${\mathfrak{b}}_n$ is not nilpotent by [L2].

Steps 1.1 and 1.2 prove the two promised properties over every field. The restriction $n\ge 2$ is sharp for this conclusion: ${\mathfrak{b}}_1=k{E}_{11}$ is one-dimensional abelian and therefore nilpotent. All witnesses are explicit and no choice is used.