Proving inequality by mathematical induction

dustbin

1. The problem statement, all variables and given/known data

I am asked to prove:
2ⁿ < (n+1)! , where n≥2

3. The attempt at a solution

Base step: set n=2, then test 2² < (2+1)!

2² = 4
(2+1)!= 3! = 3(2)(1) = 6
so 4 < 6 , which is true.

Induction hypothesis is 2^k < (k+1)!
Using this, prove 2^(k+1) < [(k+1)+1]! = (k+2)!

Attempt to solve:

starting with what I know: 2^k < (k+1)!
Multiplying both sides by 2: 2(2^k) = 2^(k+1) < 2(k+1)!

I know that 2(k+1)! < (k+2)!
since (k+2)! = (k+2)(k+1)! and because k≥2, (k+2) will be greater than 2. Thus, multiplying (k+1)! by 2 on the LHS is less than multiplying (k+1)! by (k+2) on the RHS.

Thus, since 2^(k+1) < 2(k+1)! is true, then 2^k+1 < [(k+1)+1]!.

P(k+1) follows from P(k), completing the induction step. By mathematical induction, P(n) is true for n≥2.


Thanks for any help!
EDIT: fixed a couple of type-o's.

Mark44

Welcome to PF, and very nice first post! Looks good!

dustbin

Thank you very much for your response! It was posted as an extra credit problem by my professor. I wanted to make sure my reasoning was correct before posting it on the board, as I've been having a little difficulty grasping what we covered about mathematical induction.

Mark44

It looks to me like you have the idea of induction proofs.