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Divisibility of factorials II (Posted on 2013-01-01) Difficulty: 3 of 5
Show that for all natural numbers, (n!)! is divisible by (n!)^(n-1)!.

No Solution Yet Submitted by Danish Ahmed Khan    
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Some Thoughts re: n=5 like part 1 (some thoughts) Comment 2 of 2 |
(In reply to n=5 like part 1 by Jer)


Perhaps you could apply the theory to both expressions in relation to a general prime, p.


Thus, power of p in n! =  Sum(r from 1 to inf.) of [n/pr]     (Jer/Legendre).


 - since we need only prove that for any prime, p, one expression contains a higher power of p than the other.


The problem is that these expressions contain more complicated factorials so I'm still trying, and perhaps getting closer in Case I, where the fact that

k3 1 = (k 1)(k2 + k + 1)  I feel may help us.

Edited on January 2, 2013, 2:34 pm
  Posted by Harry on 2013-01-02 14:30:49

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