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How does one prove by induction that

$n! > n^2$

for $n \geq 4$

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marked as duplicate by Martin Sleziak, Najib Idrissi, kingW3, Joel Reyes Noche, Ittay Weiss Feb 9 '15 at 12:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

This is not true when n=1 – Amr Dec 16 '12 at 20:06
How does one prove anything by induction? Can you say which part of the general method is problematic in this case? – Trevor Wilson Dec 16 '12 at 20:08
The OP probably intended for $n \geq 4$. – bzprules Dec 16 '12 at 20:10
@cardinal yes, of course! – amWhy Dec 16 '12 at 20:12
I edited it sorry it deserved a -1.... – Olivia Irving Dec 16 '12 at 20:14

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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did you mean $k^3-k-1>0$? – anegligibleperson Dec 16 '12 at 20:25
I meant $k^2-k-1>0$; I got that expression by dividing the right hand inequality by $k+1$. – bzprules Dec 16 '12 at 20:32
ahh sorry my bad – anegligibleperson Dec 16 '12 at 20:35
Thanks, I didn't see the other post answered when searching – Olivia Irving Dec 16 '12 at 23:15


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

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