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So we are asked to show that $$(p-1)(p-2)\cdots(p-r)\equiv (-1)^{r}r! \pmod{p}$$ for $r=1,2,...,p-1$. I worked on it and I want to know if my proof suffices to show what is being asked. I would also appreciate any alternative proofs.


We know that $(p-1)\equiv -1\pmod{p},\,(p-2)\equiv -2\pmod{p}, \,\,\ \cdots \,,(p-r)\equiv -r\pmod{p}$. Multiplying all of these together we have $$(p-1)(p-2)\cdots(p-r)\equiv (-1)(-2)\cdots(-r)\pmod{p}$$ which is the same $$(p-1)(p-2)\cdots(p-r)\equiv (-1)^r r!\pmod{p}$$ Q.E.D.

Yes? No? Any suggestions. Thank you in advance.

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Just fine. Nothing more needed. There is enough detail, argument is clear. – André Nicolas Jun 24 '12 at 6:54
Looks okay to me. – i. m. soloveichik Sep 22 '12 at 17:21
Your proof is good. You never use the fact that $r = 1,2,\ldots,p-1$, so it holds for any positive $r$, but if $r \geq p$, it gets really uninteresting ($r! \equiv 0 \bmod{p}$). – andybenji Dec 11 '12 at 1:11

Your proof is fine. The only suggestion that I might have is to try to formalize this a bit more by the use of induction, but everything looks correct. Whether you should use induction or not really depends on the level of the course you're taking. Have you seen induction?

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You can post the same thing as a comment right ?. But anyway its your wish . – Iyengar Jun 24 '12 at 6:26
@Iyengar I thought about that, but I'm answering the question that the OP is asking, and since answering a question in the comments is in fact disencouraged, I believe I'm following the guidelines of the site (regardless of the length of the answer and the effort that I put into it). If I'm wrong about this point, I welcome any corrections. – talmid Jun 24 '12 at 6:32
Oh, I don't know that !!. I take back my words, sorry ;) – Iyengar Jun 24 '12 at 6:33
@Iyengar Never mind :-) – talmid Jun 24 '12 at 6:36
@talmid Thank you for your input. I appreciate it. – HowardRoark Jun 24 '12 at 6:38

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