Okay, that's a different problem. To use induction to prove that P(n) is true for all [itex]n\ge a[/itex] you must prove two things:
1) That P(a) is true.
2) That if P(k) is true, for some integer [itex]k\ge a[/itex] then P(k+1) is true.
Here, P(n) is "[itex]n!> n^2[/itex]" for [itex]n\ge 4[/itex].
(1) So P(a)= P(4) is "[itex]4!> 4^2[/itex]". 4!= 4(3)(2)(1)= 24 while [itex]4^2= 16[/itex]. Yes, 24> 16.
(2) Now assume that [itex]k!> 2^k[/itex] for some k> 3. Then
[tex](k+1)!= (k+1)(k!)> (k+1)2^k[/tex]
Now we need to show that [itex](k+1)(k^2)> (k+1)^2[/itex].
Since k+1> 0 we really only need to show that [itex]k^2> k+1[/itex]. Let's do a separate induction to prove that [itex]n^2> n+ 1[/itex] for n> 1. If n= 2, [itex]2^2= 4[/itex] while [itex]2+ 1= 3[/itex]. Yes, [itex]4> 3[/itex].
Now assume that [itex]k^2> k+1[/itex] for some k. Then
[tex](k+1)^2= k^2+ 2k+ 1> (k+1)+ 2k+ 1= 3k+ 2[/tex]
[tex]= (k+1+ 1)+ 2k> (k+1+ 1)[/tex].
That is, (k+1)^2> (k+1)+ 1 so the statement "[itex]n^2> n+1[/itex]" is true for all n larger than 1.
Now that we know [itex](k+1)k^2> (k+1)^2[/itex], it follows that [itex](k+1)!> (k+1)k^2> (k+1)^2[/itex] and we are done.