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

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The discussion revolves around proving the inequality 2^n < n! for n >= 4 using mathematical induction. The user has established the base case but is uncertain about progressing from the inductive hypothesis. A suggestion is made to express n! in terms of its factors and to utilize the properties of the terms to demonstrate the inequality. The approach involves showing that if the inequality holds for k, it also holds for k+1 by manipulating the expressions appropriately. The conversation emphasizes the importance of clear steps in the induction process to complete the proof effectively.
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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
 
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NateTG helped me in the original posting, so you can disregard this (unless you want to do it for fun!) Thanks
Josh
 
joshanders_84 said:
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
 
Induction is the easiest way, I think.

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

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.
 
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