Hint number 1 :
Hint number 2 : Let denote the inverse of modulo p. Then .
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?