 Solution

Use the permitted triangularization to write $A=TU{T}^{-1}$ with $U$ upper triangular and diagonal ${\lambda }_1,\dots ,{\lambda }_n$. Products of upper-triangular matrices are upper triangular, and their diagonal entries multiply, soThe 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 chainof $n+1$ indices in $\{1,\dots ,n\}$, which is impossible. HenceThis is nilpotence from vanishing traces of powers. It proves nilpotence without substituting an unproved diagonalization for the allowed triangularization.