# A proof in number theory dealing with modular congruences.

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.

Proof:

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