January 7th, 2019, 12:33 PM  #1 
Newbie Joined: Dec 2018 From: england Posts: 7 Thanks: 0  Factorial problem 
January 7th, 2019, 01:11 PM  #2 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
$\displaystyle n(!)^k $ is divisible by $\displaystyle n!$ Now just prove that $\displaystyle n(!)^k > (n!)^{(n(!)^0 1)(n(!)^1 1)...(n(!)^{k2}1)}$ 
January 15th, 2019, 05:05 AM  #3 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
The two sides have the same elements(factors). n(!)^k has more factors , same as being greater than the left side. Proving that it can be divided by the left side is same as proving it is greater. Last edited by idontknow; January 15th, 2019 at 05:08 AM. 

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