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