1. ## Can someone please see of this induction answer makes sense??

The problem

2^n < (n+2)! for all integers n >= 0

Show for n=1

2 < 3! true

Assume for n = k, then 2^k < (k+2)!

Show for n = k+1

2^(k+1) < (k+1+2)!
(2^k)2 < (k+3)!
2(2^k) < (k+3)(k+2)!
2^k< [(k+3)/2](k+2)!

since 2^k < (k+2)!, then it is true for some multiple of (k+2)!, given that that multiple is certainly greater than 1

qed

Thank you

2. Major remark. The problem says "for all integers n >= 0", so the base case should be for n = 0.

(Pretty) minor remarks. In the beginning of the induction step, it is good to write the restriction on k, i.e., k >= 0. I guess, here the fact that 2 < k + 3 is obvious, but when I am trying to evaluate it for the first time, I immediately want to see what k is.

2^(k+1) < (k+1+2)!
(2^k)2 < (k+3)!
2(2^k) < (k+3)(k+2)!
2^k< [(k+3)/2](k+2)!
Here you start with what you need to prove and end with something similar to the induction hypothesis. This is a natural record of simplifying the inequality one has to prove. However, when proofs are written in their final form, they usually go from assumptions through intermediate statements to the final result.

3. Originally Posted by Thetheorycase
The problem

2^n < (n+2)! for all integers n >= 0

Show for n=1 see emakarov's note, n=0, 1<2 true

2 < 3! true

Assume for n = k, then 2^k < (k+2)!

Show for n = k+1

2^(k+1) < (k+1+2)!
(2^k)2 < (k+3)!
2(2^k) < (k+3)(k+2)!
2^k< [(k+3)/2](k+2)!

since 2^k < (k+2)!, then it is true for some multiple of (k+2)!, given that that multiple is certainly greater than 1

qed

Thank you
Yes,
you could also have stated that if

$2^k<(k+2)!\Rightarrow\ (2)2^k<2(k+2)!$

Then since $n$ is non-negative and $k$ is some $n$

then $(2)2^k<(k+3)(k+2)!$

if $2^k<(k+2)!$

since the smallest value of $(k+3)$ is 3.