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Re^2: A bad shuffleby tlm (Prior) 
on Mar 24, 2005 at 00:45 UTC ( #441930=note: print w/replies, xml )  Need Help?? 
N^{N1}/(N1)! is not an integer for any integer N > 2, so it is not the case that N!  N^{N}, except for N=1 and N=2. the lowliest monk Update: Here's a proof of the assertion made above. I'm sure there are better proofs of it, but this is the best I could come up with. Assume that N > 2, and let p be the largest prime in the prime factorization of N. There are three cases to consider. Suppose first that p is 2. Then, by assumption, N is a power of 2 greater than or equal to 4. Therefore, 3 is a factor of N!, and consequently N! does not divide N^{N}. Next, suppose that p > 2. If N = p^{k} for some nonnegative integer k, then N is odd and not divisible by N! whose prime factorization includes 2. This leaves the case in which p > 2, and is not the sole prime factor of N. In this case N >= 2 p. By Bertrand's postulate there exists a prime q such that p < q < 2 p <= N. Therefore, there is a factor of N!, namely q, that does not divide N^{s}, for any positive integer s. Therefore, N! does not divide N^{N}, for all N > 2.
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