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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 at 12:41

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This is not true when n=1 –  Amr Dec 16 '12 at 20:06
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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
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The OP probably intended for $n \geq 4$. –  bzprules Dec 16 '12 at 20:10
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@cardinal yes, of course! –  amWhy Dec 16 '12 at 20:12
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I edited it sorry it deserved a -1.... –  Olivia Irving Dec 16 '12 at 20:14

2 Answers 2

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

Hint:

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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