Show that for all natural numbers, (n!)! is divisible by (n!)^(n1)!.
(In reply to
n=5 like part 1 by Jer)
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/p^{r}] (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
k^{3} – 1 = (k – 1)(k^{2} + k + 1) I feel may help us.
Edited on January 2, 2013, 2:34 pm

Posted by Harry
on 20130102 14:30:49 