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Possible Duplicate:
Proof the inequality $n! \geq 2^n$ by induction

I have the following:

Prove that for all $n \in Z^+,\space n > 3 \implies 2^n < n!$

Please provide the steps and, if possible, an explanation.

Best,

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    **Hint** $\rm\displaystyle\ \frac{n!}{2^n}\ =\ \left[\frac{1}2\ \frac{2}{2}\ \frac{3}2\ \frac{4}2\right]\ \frac{5}2\: \frac{6}2\ \cdots\ \frac{n}2_{\phantom{{\frac{I}{I}}\!\!\!\!\!\!}}\ > 1\ $ since each factor is. This is a prototypical proof by [multiplicative telescopy.](http://math.stackexchange.com/search?q=user:242+telescopy)2012-04-04

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