 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 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.