Show that

(p-1)(p-2)...(p-r)==(-1)^r*r!(mod p)

for r=1, 2, ..., p-1.

I am totally lost on this one... Can you help me get started? Thanks.

Oops, this should have said Wilson's theorem in the title.

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- February 24th 2010, 06:30 AMtarheelbornFermat's Little Theorem
Show that

(p-1)(p-2)...(p-r)==(-1)^r*r!(mod p)

for r=1, 2, ..., p-1.

I am totally lost on this one... Can you help me get started? Thanks.

Oops, this should have said Wilson's theorem in the title. - February 24th 2010, 11:54 AMqmech
Remember this is all mod p.

So

1 = p+1 = 2*p+1, etc.

In particular,

p-1 = -1

p-2 = -2, etc.

Try replacing each term on the left with 'something equal to it mod p' on the right. - February 24th 2010, 12:51 PMtarheelborn
So something like

-1 * -2 * ... * -r == (-1)(-2)...(-r)(mod p)

== (-1)^r(1*28...*r)(mod p)

== (-1)^r * r! (mod p)

and we are done, right?

Thank you.