Solution
= Solution
{c}
We have $D(I_n)=\det A$. Define a function of the columns of $B$ by
$$
F(B)=D(B)+(1-\det A)\det B.
$$
It is an <alternating multilinear map>, because both summands are, and $F(I_n)=\det A+1-\det A=1$. Hence $F$ is a normalized <determinantal function>. By the asserted uniqueness of the <determinant>, $F(B)=\det B$ for every $B$. Rearranging yields
$$
\boxed{\det(AB)=\det A\det B.}
$$
This argument also covers $\det A=0$ without dividing by it.