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How can we proof that if $p$ is prime and $k$ is integer number that $1<k<p-1$ then $(p-k)!(k-1)! \equiv (-1)^k\pmod p$

user426865
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    @Andrew it's not a duplicate, he's clearly asking a completely different, although related question to the one you linked. Consider actually reading the question before you start telling the OP that he needs to look better and linking non-duplicate questions. –  Mar 20 '17 at 15:42
  • thank you very much for everyone – user426865 Mar 20 '17 at 16:14

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We see that (everything being $\mod p$) \begin{align} (p-k)!(k-1)!&\equiv(k-1)\cdot(k-2)\cdots2\cdot1\\ &\equiv(p-k)!(-1)^{k-1} (1-k)\cdot(2-k)\cdots-2\cdot-1\\ &\equiv(p-k)!(-1)^{k-1} (p-k+1)\cdot(p-k+2)\cdots(p-2)(p-1)\\ &\equiv(p-1)!(-1)^{k-1}\\ &\equiv(-1)(-1)^{k-1}\\ &\equiv(-1)^k\mod p \end{align}

Where we use Wilsons theorem at the second-last line, $(p-1)!\equiv -1\mod p$.