Proving $n! > n^3$ for all $n > a$ Prove by induction: Find a, and prove the postulate by mathematical induction.
$$\text{For all}~ n > a,~ n! > n^3$$
Where ! refers to factorial.
So far I've done a bit of it, I'll skip right to the inductive statement and assume that $k! > k^3$, then try to prove $(k+1)! > (k+1)^3$
Inductive statement: $(n+1)! > (n+1)^3$
$(n+1)^3= n^3 + 3n^2 + 3n + 1 < n! + 3n^2 + 3n + 1$ (By Induction Hypothesis)
...But now I'm stuck. Does anyone know where to go next?

Check the first number: $4! = 24 < 4^3 = 64$; $5! = 120 < 5^3 = 125$.
Now, $6! = 720 > 6^3 = 216$.
Assume that $k! > k^3$ for $k>5$. We need to prove that $(k+1)! > (k+1)^3$.
Firstly, we prove that $k^3 > (k+1)^2$. Indeed, we have $(k-1)^3 > 0$, then $k^3> 3k^2 - 3k + 1 > k^2 + 2k + 1 = (k+1)^2$.
So, we get $(k+1)! > k^3(k+1) > (k+1)^3$.
