1. Problem: Explain why $-(m+1)!(p-m-2)!\equiv(-1)^{m+1}\pmod{p}$ becomes $(m+1)!(p-m-2)!\equiv(-1)^{m+2}\pmod{p}$ and prove that for a prime $p$ and $0\le k\le p-1$ we have $k!(p-k-1)!\equiv(-1)^{k+1}\pmod{p}$. 2. Key fact: Wilson's theorem states that for a prime $p$ we have $(p-1)!\equiv -1\pmod{p}$. 3. Proof setup: Write $(p-1)! = k!\cdot (k+1)(k+2)\cdots(p-1)$. 4. Replace each factor $p-j$ with $-j$ for $1\le j\le p-k-1$ to get $(k+1)\cdots(p-1)\equiv (-1)^{p-k-1}(p-k-1)!\pmod{p}$. 5. Combine with Wilson: $-1\equiv (p-1)!\equiv k!\cdot (-1)^{p-k-1}(p-k-1)!\pmod{p}$. 6. Solve for the product: because $(-1)^{p-k-1}$ is its own inverse we obtain $k!(p-k-1)!\equiv -1\cdot (-1)^{p-k-1}=(-1)^{p-k}\pmod{p}$. 7. Parity step: since $(-1)^{p-k}=(-1)^{k+1}$ for any prime $p$ (this holds including $p=2$ by direct check) we conclude $k!(p-k-1)!\equiv (-1)^{k+1}\pmod{p}$. 8. About the sign change: if $-(m+1)!(p-m-2)!\equiv (-1)^{m+1}\pmod{p}$ then multiply both sides by $-1$ (which is invertible modulo $p$) to get $(m+1)!(p-m-2)!\equiv -(-1)^{m+1}=(-1)^{m+2}\pmod{p}$. 9. Final answer: for prime $p$ and $0\le k\le p-1$ we have $k!(p-k-1)!\equiv (-1)^{k+1}\pmod{p}$.