= Solution
{c}
If $A$ is a <strictly upper triangular matrix>, then $A_{ij}=0$ unless $j-i\ge1$. We prove by <mathematical induction> that
$$
(A^k)_{ij}=0\qquad\text{whenever }j-i<k.
$$
The assertion for $k=1$ is the definition. In the <matrix multiplication> formula
$$
(A^{k+1})_{ij}=\sum_{\ell=1}^n(A^k)_{i\ell}A_{\ell j},
$$
a nonzero summand would require $\ell-i\ge k$ and $j-\ell\ge1$, hence $j-i\ge k+1$. This proves the induction step. Since $j-i\le n-1$ for all entries, $\boxed{A^n=0}$. In particular, every <strictly upper triangular matrix> is a <nilpotent matrix>.
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