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How does one prove by induction that

$n! > n^2$

for $n \geq 4$

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    See also: [Prove by induction that n^2](http://math.stackexchange.com/questions/1140396/prove-by-induction-that-n2n) and [Hint in Proving that $n^2\le n!$](http://math.stackexchange.com/questions/764808/hint-in-proving-that-n2-le-n)2015-02-09

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To prove that this inequality holds for $n \geq 4$, first we verify the base case, which is trivial, as $24 >16$.

Now assume for some $k$ that $k! > k^2$. Then $(k+1)! > (k+1)k^2 = k^3+k^2 > (k+1)^2$. We can verify the right hand inequality, as this implies that $k^2 > k+1 \implies k^2-k-1>0$, which is clearly true for $k \geq 4$; the inductive step has thus been proven and we're done.

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    Thanks, I didn't see the other post answered when searching2012-12-16
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Hint:

If $n! > n^2$ holds, can you show that $ (n+1)! = n! (n+1) > (n+1)^2 = n^2 +2 n+1?$