= Solution
For an $n\times n$ <matrix> $A$, the characteristic <polynomial> is
$$
\chi_A(t)=\det(tI-A).
$$
The <Cayley-Hamilton theorem> states that $\chi_A(A)=0$.
Over $\mathbb C$, choose a <basis> in which $A$ is upper triangular, with diagonal entries $\lambda_1,\ldots,\lambda_n$. For the standard invariant flag $V_j=\langle e_1,\ldots,e_j\rangle$,
$$
(A-\lambda_jI)V_j\subseteq V_{j-1}.
$$
The factors $A-\lambda_jI$ commute, so applying their product in descending order sends $V_n$ successively into $V_{n-1},\ldots,V_0=0$. Hence
$$
0=\prod_{j=1}^n(A-\lambda_jI)=\chi_A(A),
$$
which proves the theorem.
Direct expansion gives the commutator product rule:
$$
\begin{aligned}
[X,YZ]
&=XYZ-YZX\\
&=(XY-YX)Z+Y(XZ-ZX)\\
&=[X,Y]Z+Y[X,Z].
\end{aligned}
$$
Put $C=[B,A]$. Since $C$ commutes with $A$, repeated use of the product rule gives
$$
[B,A^r]=\sum_{j=0}^{r-1}A^jCA^{r-1-j}=rA^{r-1}C.
$$
By <linearity>, for every <polynomial> $\varphi$,
$$
[B,\varphi(A)]=\varphi'(A)C.
$$
Let $D(X)=[B,X]$ and suppose $f(A)=0$. For $k=1$,
$$
f'(A)C=D(f(A))=0.
$$
Assume inductively that
$$
uC^m=0,
\qquad
u=f^{(k)}(A),\quad m=2^k-1.
$$
Both $u$ and $C$ are <polynomials> in, or commute with, $A$, so $uC^m=C^mu=0$. Apply the derivation $D$ to $uC^m=0$ and multiply on the left by $C^m$:
$$
0=C^mD(u)C^m+C^muD(C^m)=C^mD(u)C^m.
$$
Since $D(u)=f^{(k+1)}(A)C$, this says
$$
f^{(k+1)}(A)C^{2m+1}
=f^{(k+1)}(A)C^{2^{k+1}-1}=0.
$$
The induction is complete. Taking $f=\chi_A$ and $k=n$ gives
$$
n!\,C^{2^n-1}=0.
$$
\b[Thus $[B,A]$ is nilpotent, which is the <Jacobson lemma for a commuting commutator>.]
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