if $p\in\mathbb P,r\in\mathbb N^*$and$r<p$, $(-1)^r r!\equiv-1(\mod p)$,show that $(p-r+1)!\equiv-1(\mod p)$.
My answer：Let's say $p$ is an odd prime,then
$$-1\equiv(p-1)!=r!(r+1)\cdots(p-1)\equiv(-1)^{p-r-1}r!(p-r-1)!\\ \equiv-(p-r-1)!\equiv(\mod p) $$ so $$(p-r+1)!\equiv(p-r)(p-r+1)!\equiv r(r-1)(\mod p)$$ We just have to prove $r(r-1)\equiv-1(\mod p)$. What to do next？
