7
$\begingroup$

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.

  • 0
    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 ($r! \equiv 0 \bmod{p}$).2012-12-11

1 Answers 1

1

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?

  • 1
    @talmid Thank you $f$or your input. I appreciate it.2012-06-24