= Solution
Since $31$ is a <prime number>, the <Wilson theorem> states that $30!\equiv-1\pmod{31}$. But $30!\equiv(-1)(-2)28!=2\cdot28!$, so the <modular inverse> of $2$, which is $16$, gives
$$
28!\equiv-16\equiv15\pmod{31}.
$$
The <Fermat little theorem> states that $a^{p-1}\equiv1\pmod p$ when $p$ is prime and $p\nmid a$. Therefore $13^{30}\equiv1\pmod{31}$, and
$$
13^{28}\equiv(13^2)^{-1}\equiv14^{-1}\equiv20\pmod{31},
$$
since $14\cdot20=280=9\cdot31+1$. Combining these <modular congruences>,
$$
28!13^{28}\equiv15\cdot20=300\equiv\boxed{21}\pmod{31}.
$$
This representative lies between $0$ and $30$, so it is the least nonnegative residue.
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