= Solution
Write $B=D+K$, where $D$ has $A$ in each diagonal block and zero elsewhere, while $K$ has $I$ immediately above the diagonal and zero elsewhere. These <block matrices> satisfy $DK=KD$: either product has $A$ immediately above the diagonal. Also, $K^i$ has $I$ on its $i$th superdiagonal and is zero elsewhere.
The <binomial theorem for commuting matrices> therefore yields
$$
B^j=\sum_{r=0}^j\binom jrD^{j-r}K^r.
$$
Only $r=i$ can contribute to block $(k,k+i)$, provided that block lies in the <matrix>. Hence
$$
\boxed{(B^j)_{k,k+i}=\begin{cases}
0,&j<i,\\
\binom ji A^{j-i},&j\geq i.
\end{cases}}
$$
Thus $\beta_{i,j}=\binom ji$, interpreted as an element of the <field>. The formula remains valid in positive <field characteristic>: some of these coefficients can then vanish. This possibility matters in the next part.
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