Use M.I. to prove that n! > n^3 for n > 5
The Attempt at a Solution
I already proved n! > n^2 for n>4, but this is nothing like that.
This is my inductive step so far.
(k+1)! > (k+1)^3
(k+1)! - (k+1)^3 > 0[/B]
(k+1)! - (k+1)^3 = (k+1)[k! - (k+1)^2]
> (k+1)[k^3 - (k+1)^2]
This is how far I can get, any hints? Am I going in the right track or am I going in the wrong direction completely?