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 (Rofl)