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# Prove that n! does not divide n^n for n>2 Watch

1. 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.
2. Hmmmm

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

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

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

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!
4. (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
5. I'd go for a direct proof, rather than induction.

Spoiler:
Show

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

Spoiler:
Show

Find a factor of n! that is coprime to n.
Spoiler:
Show

n-1 is clearly coprime to n. But why does this mean it can't divide n^n?
8. (Original post by DeeperBlue)
Spoiler:
Show

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).
9. 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!
10. (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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Updated: April 12, 2013
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