= Solution
Use <induction> on $\dim V$. By part (b), there is a one-dimensional invariant <linear subspace> $U=Kv$ on which $L$ acts as zero. The induced operators on $V/U$ remain nilpotent, so <induction> gives a simultaneous strictly upper triangular <basis> there. Part (c) lifts it together with $v$ to such a <basis> of $V$.
Thus all elements strictly lower one common complete flag. A nested <Lie bracket> of sufficiently many flag-lowering operators lowers it past zero. Consequently the <lower central series> of $L$ terminates, proving that \b[the <Lie algebra> is nilpotent]. In fact its nilpotency class is at most $n-1$ for $n>1$; the zero algebra covers the one-dimensional case.
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