Question: Let $p$ be prime, and suppose that $r$ is a positive integer less than $p$ such that $(−1)^rr! ≡ −1($mod $ p)$. Show that $(p − r + 1)! ≡ −1 ($mod $ p)$.
I'm not really sure what to do. My first strategy was to take $(−1)^rr!$, and expand it out. Doing so, we can get: $(-r)(-r+1)(-r+2)...(-2)(-1)$ My next assumption was to just add $p$ to all of these terms, but that doesn't get us what we want. Ultimately, it seems like we want to show $(p-r+1)(p-r)(p-r-1)(p-r-2)(2)(1)\equiv(−1)^rr!$ . There are more terms though potentially in $(p-r+1)!$ than $(−1)^rr!$, so I'm not really sure what to do. Perhaps connecting Wilson's theorem with the given? Hints would be greatly appreciated. Thanks!
Edit: Here's a screenshot directly from the book I am using.
The book is Elementary Number Theory, 6th edition by Kenneth H. Rosen.
Are you sure that what you are trying to prove is even true? Below you can find a counterexample to your problem.
Let $p$ be a prime and let $r=1$. Then it is always the case that $(-1)^rr!\equiv-1\bmod p$, but $(p-r+1)!\equiv p!\equiv 0 \bmod p$.
Just in case anyone wants a second counterexample, since $r=1$ feels kind of silly...
Let $p$ be a prime greater than $3$, and let $r=p-1$. Then it is always the case that $(-1)^rr!\equiv(p-1)!\equiv-1\bmod p$ by Wilson's Theorem, but $(p-r+1)!\equiv 2!\equiv 2 \bmod p$.