= Solution
Each entry of $\operatorname{adj}(M-tI)$ is, up to sign, the determinant of an $(n-1)\times(n-1)$ minor whose entries are affine polynomials in $t$. The <Leibniz formula for determinants> therefore makes each adjugate entry a polynomial in $t$ of degree at most $n-1$.
For arbitrary $A,B$, choose positive sequences $s_j,t_j\to0$ for which $A-s_jI$ and $B-t_jI$ are nonsingular. The nonsingular identity from part (b) applies to their product:
$$
\operatorname{adj}((A-s_jI)(B-t_jI))
=\operatorname{adj}(B-t_jI)\operatorname{adj}(A-s_jI).
$$
Every entry is a polynomial, hence continuous, in the matrix entries. Letting $j\to\infty$ proves
$$
\boxed{\operatorname{adj}(AB)
=\operatorname{adj}(B)\operatorname{adj}(A)}
$$
for all square matrices.
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