Originally Posted by

**cassiopeia1289** So - I know all the steps, I just need help with a gap.

Prove $\displaystyle n! > n^2 \exists n \geq4$

Ok, proof by Induction:

Since 4! = 24 which is $\displaystyle >(4)^2 = 16$, $\displaystyle n!>n^2$ for n=4.

Let k be a positive integer [math\geq4[/tex] and suppose $\displaystyle k!>k^2$.

Then, $\displaystyle k!(k+1) > k^2(k+1)$

$\displaystyle (k+1)! > k^2(k+1) = k^3+k^2$

Here's where I'm stuck. I know $\displaystyle k\geq4$ but how can I show that $\displaystyle (k+1)^2 < k^3+k^2$?

From then, I can show $\displaystyle (k+1)! > k^3+k^2 > (k+1)^2$

Therefore, $\displaystyle (k+1)! > (k+1)^2$, so $\displaystyle n! > n^2 for n \geq4$.

Please help me out with this way of solving the proof, I'm sure there are other ways.

any help: please!!!??