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I have to prove $n<n!$ for all $n>2$ by mathematical induction.

I did it as follows. I proved the base case.

Then let it be true for $K>2$:

$$ K<K! $$

I have to prove,

$$ K+1<(K+1)! $$

$$ K<K! $$

Adding $1$ on both sides

$$ K+1<K!+1<(K+1)! $$

Hence

$$ K+1<(K+1)! $$

Is the last step I did ($K!+1<(K+1)!$) Right?

Please help me out. Thanks!

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I think you should prove $K!+1<(K+1)!$ which is not so hard. –  the symplectic camel Nov 5 '11 at 18:53
    
No. For the n=3 case, you just check if 3<3! which is true. Then assume for n=k case and prove for the n=k+1 case. –  the symplectic camel Nov 5 '11 at 19:02
    
@KaratugOzanBircan: Thanks –  Fahad Uddin Nov 5 '11 at 19:06
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What do you mean by n2? –  Rankeya Nov 5 '11 at 19:17
    
@Rankeya: I guess, it should be $n>2.$ –  the symplectic camel Nov 5 '11 at 19:32

3 Answers 3

up vote 9 down vote accepted

Why use induction at all when you can see plainly that $n! = n\cdot (n-1) \cdots 1 \ge n\cdot (n-1) > n$ if $n>2$ ?

If this is an exercise, it's a silly one. It's exercises like this that give a bad name to induction...

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2  
Amen. Though to be fair, exercises that specify an unreasonable method are hardly limited to induction, never mind exercises that specify a reasonable method when others would do as well. –  Brian M. Scott Nov 5 '11 at 21:07
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or similarly $n! = n\cdots 2\cdot 1 \ge n\cdot 2 \gt n$ if $n \gt 2$, making it clearer why $2$ matters. –  Henry Nov 6 '11 at 0:45
    
@Henry, sure. It proves something stronger than required but so does my argument, and mine provides a quadratic lower bound :-) –  lhf Nov 6 '11 at 2:12

Starting with $K \lt K!$,

multiplying both sides by $K+1 \gt 0$ you have $K(K+1) \lt (K+1)!$,

but since $K\gt 2$ you have $K+1 \lt K(K+1) $ and so $K+1 \lt (K+1)!$

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Since in your case $K > 2$, the statement $K!+1 < (K+1)!$ is true. This is because $(K+1)! = K!(K+1) = K!K + K!$, and $K!K > 1$ for $K > 1$.

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