given that $(p-1)! \equiv -1 \mod{p}$, I was able to find a formula for $(p-a)! \mod{p}$ for $a \in \mathbb{Z}$:

$(p-1)! = (p-1)(p-2)! \equiv (-1)(p-2)! \mod{p} \equiv -1 \mod{p} \Rightarrow$

$(p-2)! \equiv 1 \mod{p}$

Using induction, one can show for $a>0$,

$(a-1)!(p-a)! \equiv (-1)^a \mod{p}$

so if $a$ is even, $(p-a)!$ is the multiplicative inverse of $(a-1)!$ modulo $p$

and if $a$ is odd, $-(p-a)!$ is the multiplicative inverse of $(a-1)!$ modulo $p \Rightarrow$

$(-1)^a(p-a)!$ is the multiplicative inverse of $(a-1)!$ modulo $p$ if $a>0$

Now when $a \leq 0, (p-a)! \equiv 0 \mod{p}$

This is because $p! \equiv 0 \mod{p}$, so for $k>0$,

$(p+k)! = p!(p+1) \dots (p+k) \equiv 0*(p+1) \dots (p+k) \mod{p} \equiv 0 \mod{p}$