Derived algebra of a solvable linear Lie algebra is nilpotent

Suppose first that $n=\dim V>0$. By [L1], choose a basis of $V$ in which every element of $\mathfrak{g}$ is upper triangular. The diagonal of $AB-BA$ is zero for upper triangular $A,B$, so every bracket is strictly upper triangular. Since strictly upper triangular matrices form a linear subspace, every element of $[\mathfrak{g},\mathfrak{g}]$, which is a finite linear combination of brackets, is strictly upper triangular and therefore has its $n$th power zero.

Put $\mathfrak{h}=[\mathfrak{g},\mathfrak{g}]$. It is a finite-dimensional Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$, and step 1.1 says that for each $x\in \mathfrak{h}$ some $m>0$ satisfies $x^m=0$. On $\mathrm{End}(V)$ one has ${\mathrm{ad}}_x=L_x-R_x$; the operators $L_x$ and $R_x$ commute, and every term of $(L_x-R_x{)}^{2m-1}$ contains either $L_x^m$ or $R_x^m$. Hence ${\mathrm{ad}}_x$ is nilpotent on $\mathrm{End}(V)$ and therefore on its invariant subspace $\mathfrak{h}$. Engel's theorem [L2] makes $\mathfrak{h}$ nilpotent. If $V=0$, then $\mathfrak{g}=0$ and both conclusions are immediate.