Proving 2^n < n! using Induction: Where to Start?

[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

 

[joshanders_84]

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

 

[Andrew Mason]

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

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

 

[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 &lt; 2.k! &lt; (k+1).k! = (k+1)! because (k+1) &gt; 2 for all k &gt; 1.

And you're done.