$(p\!-\!1\!-\!h)!\,h! \equiv (-1)^{h+1}\!\!\pmod{\! p}\,$ [Wilson Reflection Formula]

Hint $\ $ Wilson's theorem implies that any complete system of representatives of nonzero remainders mod $\,p\,$ has product $\equiv -1.\,$ In particular this is true for any sequence of $\,p\,$ consecutive integers, after we remove its unique $\rm\color{#c00}{multiple}$ of $\,p.\,$ Your special case is the sequence $$\, -h,\,-h\!+\!1,\ldots,-1,\require{cancel}\color{#c00}{\cancel{0,}} 1,2,\ldots, k\ \ \ \text{whose product is}\,\ \ (-1)^h h!\,k!\equiv -1$$

Remark $\ $ This is slight reformulation of the Wilson reflection formula mentioned yesterday

$$ k! = (p\!-\!1\!-\!h)! \equiv \frac{(-1)^{h+1}}{h!}\!\!\pmod{\! p},\,\ \ 0\le h< p\ {\rm prime}\qquad $$

Use the fact that

$$h! = (-1)^h (p-1)(p-2) \dots (p-h) \mod p$$