# Induction proof of 2^n < n!

1. Jan 11, 2005

### joshanders_84

I posted this elsewhere but realized it ought to be in the homework section. I have to use induction to prove that for n>=4, 2^n < n! is true, but I don't know wehre to start. I have the base case proven, but then I don't know where to go after I have my Inductive Hypothesis that it works for all n's greater than 4. Any help would be very appreciated. Thank you
Josh

2. Jan 11, 2005

### joshanders_84

NateTG helped me in the original posting, so you can disregard this (unless you want to do it for fun!) Thanks
Josh

3. Jan 11, 2005

### Andrew Mason

Write out the expression for n!

$$n! = n(n-1)(n-2)(n-3)...(n-(n-1))$$

Then subtract each term by a positive number so that each term in the product is equal to 2.

AM

4. Jan 11, 2005

### Curious3141

Induction is the easiest way, I think.

EDIT : Sorry, I just saw the title of your post, and you wanted the induction proof.

The initial verification for the case of 4 is easy. Say the inequality holds for some $k$.

Then $$2^{k+1} = 2.2^k < 2.k! < (k+1).k! = (k+1)!$$ because $(k+1) > 2$ for all $k > 1$.

And you're done.

Last edited: Jan 11, 2005