# Thread: prove that the binomial (p-1) over k is congruent to -1^k modulo p

1. ## prove that the binomial (p-1) over k is congruent to -1^k modulo p

if p is an odd prime (i.e. greater than 2) and k is an integer such that 1 is less than or equal to k is less than or equal to p-1 then:

the binomial of p-1 over k is congruent to (-1)^k modulo p.

"binomial of" means: p-1 over k = (p-1)!/(k![(p-1)-k]!)

i know that this reduces to:

(p-1)(p-2)...(p-k) over 1(2)(3)...k

it is not clear to me how this works though. I realize that all the elements of (p-1)! occur in the binomial, whether they be in the numerator or denominator, as in the example p=11 and k=5, where:

10(9)(8)(7)(6)/[1(2)(3)(4)(5)]

this above expression contains all of (p-1)!

how do i prove the first statement is true?

2. Hint number 1 : $\displaystyle (p-1)! \equiv -1 \mod p$

Hint number 2 : Let $\displaystyle k^{-1}$ denote the inverse of $\displaystyle k$ modulo p. Then $\displaystyle k^{-1}(p-k) \equiv -1 \mod p$.