= Solution
The <Wilson theorem> states that for a prime $p$,
$$
(p-1)!\equiv-1\pmod p.
$$
Indeed, every nonzero residue modulo $p$ has a unique multiplicative inverse. The only residues equal to their own inverses solve $x^2\equiv1\pmod p$, hence are $1$ and $-1$. Pairing every other residue with its distinct inverse leaves
$$
(p-1)!\equiv1\cdot(-1)\equiv-1\pmod p.
$$
The <Fermat little theorem> states that for prime $p$,
$$
a^p\equiv a\pmod p;
$$
equivalently, if $p\nmid a$, then $a^{p-1}\equiv1\pmod p$.
Wilson's theorem at $p=31$ gives
$$
30!=30\cdot29\cdot28!\equiv2\cdot28!\equiv-1\pmod{31},
$$
so $28!\equiv15\pmod{31}$. Also $729\equiv16\pmod{31}$, and therefore
$$
\boxed{28!\cdot729\equiv15\cdot16\equiv23\pmod{31}}.
$$
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