Solution
= Solution
The <Nilpotent Jordan block> $J_d$ shifts each standard basis vector one place toward the first coordinate. Therefore
$$
(J_d^n)_{ij}
=\begin{cases}
1,&j-i=n,\\
0,&\text{otherwise}.
\end{cases}
$$
Thus $J_d^0=I$, for $1\leq n<d$ the matrix has ones precisely on its $n$th superdiagonal, and
$$
\boxed{J_d^n=0\quad\text{for }n\geq d}.
$$