 Solution

Suppose $S^b=0$. Use the increasing chain of vector subspacesChoose a basis successively, extending a basis of each kernel of a linear map to one of the next kernel of a linear map. If a basis vector first enters at level $j$, its image under $S$ belongs to $\ker S^{j-1}$, spanned entirely by earlier basis vectors. Thus every matrix column has nonzero entries only above its diagonal: the matrix is strictly upper triangular.

Conversely, for such a matrix and its ordered basis $e_1,\dots ,e_n$, one has $S⟨e_1,\dots ,e_j⟩\subseteq ⟨e_1,\dots ,e_{j-1}⟩$. Iterating $n$ times kills every vector. Hence $\boxed{S^n=0}$, establishing both directions without assuming a Jordan form.