= Solution
Use the permitted triangularization to write $A=TUT^{-1}$ with $U$ upper triangular and diagonal $\lambda_1,\ldots,\lambda_n$. Products of upper-triangular <matrices> are upper triangular, and their diagonal entries multiply, so
$$
0=\operatorname{tr}(A^k)=\operatorname{tr}(U^k)=\sum_{j=1}^n\lambda_j^k
\qquad(1\le k\le n).
$$
The result of part b makes every $\lambda_j$ zero. Thus $U$ is strictly upper triangular. To see directly that $U^n=0$, a potentially nonzero entry in a product of $n$ factors requires a chain
$$
i=i_0<i_1<\cdots<i_n=j
$$
of $n+1$ indices in $\{1,\ldots,n\}$, which is impossible. Hence
$$
\boxed{A^n=TU^nT^{-1}=0.}
$$
This is <nilpotence from vanishing traces of powers>. It proves nilpotence without substituting an unproved diagonalization for the allowed triangularization.
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