= Solution
{c}
First apply the <commutator product rule> in the form $[AB,C]=A[B,C]+[A,C]B$:
$$
[N,a_j^*]=\sum_{k=1}^2\left(a_k^*[a_k,a_j^*]+[a_k^*,a_j^*]a_k\right)=a_j^*.
$$
Repeated use of the <commutator product rule> gives, for $n\geq1$,
$$
[N,(a_j^*)^n]=\sum_{r=0}^{n-1}(a_j^*)^r[N,a_j^*](a_j^*)^{n-1-r}
=n(a_j^*)^n.
$$
This is the <commutator with powers of an eigenoperator>. For $n=0$, the <commutator> with the <identity operator> is zero, agreeing with the formula.
Each $a_k^*a_k$ is unchanged by taking the <adjoint operator>, so $N^*=N$ in this algebra. Taking the <adjoint of a commutator> yields $[N,(a_j^*)^n]^*=[a_j^n,N]=na_j^n$. By antisymmetry,
$$
\boxed{[N,(a_j^*)^n]=n(a_j^*)^n,\qquad [N,a_j^n]=-na_j^n,\qquad n=0,1,2,\ldots.}
$$
In particular, each <creation operator> raises the <number operator> <eigenvalue> by one, and each <annihilation operator> lowers it by one.
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