Use math induction to show that
1) n!>= n^2 for n>=6
2) n!>=n^3 for n>=6
I will assume that you have done the base case and that we assume this is true for some $\displaystyle k \ge 6$
So we need to show that:
$\displaystyle (k+1)! \ge (k+1)^2$,
well:
$\displaystyle (k+1)! = k!~ k$
and by supposition $\displaystyle k! \ge k^2$, so:
$\displaystyle (k+1)! = k! k \ge k^3$
So if we can prove that $\displaystyle k^3 \ge (k+1)^2$ we are done, but $\displaystyle k^3 \ge (k+1)^2$ is equvalent to $\displaystyle k^3-k^2-2k-1 \ge 0$. So to compete the proof of the induction step you need to show that:
$\displaystyle k^3-k^2-2k-1 = 0$
has no real root greater than or equal to $\displaystyle 6$ (this is sufficient as for $\displaystyle k$ greater than its largest real root the cubic is positive).
This you do by making the substitution $\displaystyle x=k-6$, then using Descartes rule of signs to show that this has no positive roots.
(In fact the only real root of $\displaystyle k^3-k^2-2k-1 = 0$ is near $\displaystyle k=2.1$)
RonL