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November 29th, 2017, 09:03 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 181 Thanks: 23  Another factorial inequality
Prove that $\displaystyle n!^2 \geq n^n$ for $\displaystyle n\in N$

November 29th, 2017, 02:45 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,397 Thanks: 546 
Mathematical induction can work. For n=1 or 2, equality. For n=3, 36 > 27. In general. $(n+1)!^2=(n+1)^2(n!)^2$ $(n+1)^{n+1}=(n+1)n^n(1+\frac{1}{n})^n<(n+1)n^ne<( n+1)^2(n!)^2$, for n > 2. 

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