= Solution
Suppose $S^b=0$. Use the increasing chain of <vector subspaces>
$$
0=\ker S^0\subseteq\ker S\subseteq\ker S^2\subseteq\cdots\subseteq\ker S^b=V.
$$
Choose 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 matrix>[strictly upper triangular].
Conversely, for such a <matrix> and its ordered <basis> $e_1,\ldots,e_n$, one has $S\langle e_1,\ldots,e_j\rangle\subseteq\langle e_1,\ldots,e_{j-1}\rangle$. Iterating $n$ times kills every <vector>. Hence $\boxed{S^n=0}$, establishing both directions without assuming a Jordan form.
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