# Thread: Trying to prove that 2^n < n! for n>= 4

1. ## Trying to prove that 2^n < n! for n>= 4

Is my proof correct?

2^n < n! for all n >= 4
Step 1: solve for base case
2^4 < 4! --> 16 < 24

Assume p(n) is true i.e. 2^n < n! is true

p(n+1)
2^(n+1) < (n+1)!

Try to solve through inductive reasoning.
In this case, just multiply both sides by (n+1)
2^n (n+1) < (n!)(n+1)

Since
n>1 we can say 2^n(n+1) > (2)2^n
(2)2^n = 2^(n+1)

So:
(n+1)! = n!(n+1) > (2^n)(n+1) > (2)2^n = 2^(n+1)
which means
2^(n+1) < (n+1)!

2. Originally Posted by Bruins1
Is my proof correct?

2^n < n! for all n >= 4
Step 1: solve for base case
2^4 < 4! --> 16 < 24

Assume p(n) is true i.e. 2^n < n! is true

p(n+1)
2^(n+1) < (n+1)!

Try to solve through inductive reasoning.
In this case, just multiply both sides by (n+1)
2^n (n+1) < (n!)(n+1)

Since
n>1 we can say 2^n(n+1) > (2)2^n
(2)2^n = 2^(n+1)

So:
(n+1)! = n!(n+1) > (2^n)(n+1) > (2)2^n = 2^(n+1)
which means
2^(n+1) < (n+1)!
Yes ,i think so

<, n>, prove