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    It seems to me that this is best proved by induction, but I get stuck on the inductive step.

    Suppose n! doesn't divide n^n , but that (n+1)! divides (n+1)^(n+1). Then for some k, k(n+1)! = (n+1)^(n+1) so that kn! = (n+1)^n

    I need to now get a contradiction but now showing that n! does divide n^n but I can't see how. I already tried using the binomial expansion on the RHS. Any suggestions would be appreciated.
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    Hmmmm

    \dfrac{3!}{3^3} = \dfrac{3*2*1}{3*3*3} = \dfrac{2*1}{3*3}

    My problem is that this is so obvious that a proof seems unnecessary

    \dfrac{n!}{n^n} = \dfrac{(n-1)!}{n^{n-1}}

    And, since (n-1)! is the product of n-1 numbers each less than n the second fraction cannot possibly work
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    (Original post by TenOfThem)
    Hmmmm

    \dfrac{3!}{3^3} = \dfrac{3*2*1}{3*3*3} = \dfrac{2*1}{3*3}

    My problem is that this is so obvious that a proof seems unnecessary

    \dfrac{n!}{n^n} = \dfrac{(n-1)!}{n^{n-1}}

    And, since (n-1)! is the product of n-1 numbers each less than n the second fraction cannot possibly work
    I am trying to show that n! does not divide n^n, not that n^n does not divide n!
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    (Original post by DeeperBlue)
    I am trying to show that n! does not divide n^n, not that n^n does not divide n!
    Use of the word "into" may have made things clearer
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    I'd go for a direct proof, rather than induction.

    Spoiler:
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    Find a factor of n! that is coprime to n.
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    Surely you can just say that n-1 does not divide n for n>2, then ust check the cases n=1,2 yourself?
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    (Original post by ghostwalker)
    I'd go for a direct proof, rather than induction.

    Spoiler:
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    Find a factor of n! that is coprime to n.
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    n-1 is clearly coprime to n. But why does this mean it can't divide n^n?
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    (Original post by DeeperBlue)
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    n-1 is clearly coprime to n. But why does this mean it can't divide n^n?
    Because if it has no (prime) factors in common with n, then it has no (prime) factors in common with n^n (which has the same prime factors as n).
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    well assume it does divide n and then get a contradtion.

    the contradiction may be along the lines off:

    n^n / n!= m * n! + r , where m is an integer and r is a remain err 0 <r <n!
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    (Original post by ghostwalker)
    Because if it has no (prime) factors in common with n, then it has no (prime) factors in common with n^n (which has the same prime factors as n).
    Thanks, this is exactly what I was missing.
 
 
 
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