I need to use mathematical induction to prove $\displaystyle 2^n<n!$ for all n>=4. The base case is easy, and in the inductive step I get to $\displaystyle 2(2^n)<(n+1)n!$ But am I going in the wrong direction, or am I just missing something?

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- Jul 30th 2007, 10:15 AMtottencmathematical induction again
I need to use mathematical induction to prove $\displaystyle 2^n<n!$ for all n>=4. The base case is easy, and in the inductive step I get to $\displaystyle 2(2^n)<(n+1)n!$ But am I going in the wrong direction, or am I just missing something?

- Jul 30th 2007, 10:41 AMtukeywilliams
Assume $\displaystyle 2^k < k! $ for $\displaystyle k \geq 4 $. Then $\displaystyle 2^{k+1} < 2k! < (k+1)! = k!(k+1) $. $\displaystyle k+1 > 2 $ for $\displaystyle k >1 $. Hence $\displaystyle 2^{k+1} < (k+1)! $ for $\displaystyle k \geq 4 $ and we have proved the inductive step.

- Jul 30th 2007, 10:49 AMtottenc
Sorry, stupid typo

$\displaystyle 2^n<n!

$ for all $\displaystyle n>=4$ - Jul 30th 2007, 11:16 AMPlato
Assume that $\displaystyle N \ge 4\quad \& \quad 2^N < N!$ for the inductive step.

Then $\displaystyle 2^{N + 1} = 2\left( {2^N } \right) < \left( {N + 1} \right)\left( {N!} \right) = \left( {N + 1} \right)!$ - Jul 30th 2007, 05:14 PMtottenc
The answer to these things is always staring me right in the face, and I just can't see it until someone points it out. Thanks again

- Jul 31st 2007, 04:34 AMtopsquark