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

  1. #1
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    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?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Hint number 1 : (p-1)! \equiv -1 \mod p

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