Proving n^n > 2^n *n using the Binomial theorem

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SUMMARY

The discussion centers on proving the inequality n^n > 2^n * n! for n > 6 using the Binomial theorem. The Binomial theorem is defined as (x+y)^n = ∑_{k=0}^n (n choose k) x^{n-k} y^k. A participant successfully proved the Binomial theorem via induction but struggled to apply it to the inequality. They attempted to manipulate the equation by setting n = (x+y) and x+y = 2, but these approaches did not yield a solution.

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Homework Statement


Prove that n^n > 2^n * n! when n > 6 using the Binomial theorem.
I just proved the Binomial theorem using induction which was not that difficult but in applying what I learned through it's proof I am having difficulty.

Homework Equations


Binomial theorem = (x+y)^n = \sum_{k=0}^n\binom{n}{k}x^{n-k}y^k


The Attempt at a Solution


I attempted setting n= (x+y) to convert the left side of the equation into the form of the binomial theorem, as well as turning the right hand side into the form of the binomial theorem by setting x+y = 2 both to no avail. Actually the "closest" (I put this in quotes because as I couldn't solve it, I have no idea how close I really was) I got was by using induction and turning the equation into {\frac{(n+1)^n }{2}}= 2^n + n!
Thanks for the help guys.
 
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Actually, the last equation I wrote have n!/2
thanks
 

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