Yes ,i think soIs 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)!