= Solution
{c}
The common <Fock vacuum> obeys $N\psi=\sum_j a_j^*(a_j\psi)=0$. Put $B=(a_1^*)^{n_1}(a_2^*)^{n_2}$. By the <commutator product rule> and part (a),
$$
[N,B]=n_1B+n_2B=(n_1+n_2)B.
$$
Moving $N$ through $B$ therefore gives
$$
NB\psi=BN\psi+[N,B]\psi=(n_1+n_2)B\psi.
$$
Dividing by $\sqrt{n_1!n_2!}$ proves
$$
\boxed{N\psi_{n_1,n_2}=(n_1+n_2)\psi_{n_1,n_2}.}
$$
The constructed <number state> is nonzero: the relation $a_j(a_j^*)^r\psi=r(a_j^*)^{r-1}\psi$ follows from $[a_j,a_j^*]=I$ and $a_j\psi=0$. Commuting different modes and successively moving each <creation operator> across the <inner product> yields
$$
\|B\psi\|^2=n_1!n_2!\,\|\psi\|^2=n_1!n_2!.
$$
Thus the given factorial denominator also gives unit <norm>, as expected for a <number state>.
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