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?
The induction hypothesis/statement lets you assume that $k!>k^3$ up to some $k$, not that $(k+1)!>(k+1)^3$ for said $k$. After you expand $(k+1)^3$ you are correct that $$k^3+3k^2+3k+1<k!+3k^2+3k+1$$ Now is where you should invoke the value of $a$. I'll let you find it still but what I can tell you is that $k>3$. So $$\begin{align}k!+3k^2+3k+1 &< k!+(k)k^2+(k^2)k+(k^3)\cdot 1 \\ &= k! + 3k^3 \\ &<k! + k\cdot k^3 \\ &<k! + k\cdot k! \tag{Induction Hypothesis} \\ &= k!(k+1) \\ &= (k+1)! \end{align}$$
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 $(k+1)! > k^3(k+1) > (k+1)^3$.
Claim: $n!>n^3$ for all $n\geq 6$
Base case: For $n=6$ we have $n!=720>216=6^3$
Let us assume for our induction hypothesis that there is some $k\geq 6$ such that $k!>k^3$. We want to show that it is also true for $k+1$.
(here you should not begin with $(k+1)!>k^3$ and try to reach a tautology. This should instead be the final line/conclusion)
$(k+1)! = (k+1)(k!)>^{\text{I.H.}}(k+1)k^3 >^{(\dagger)} (4)k^3 =k^3+k^3+k^3+k^3> k^3+3k^2+3k+1=(k+1)^3$
where $\text{I.H.}$ denotes where we applied the induction hypothesis and the step at $(\dagger)$ is because $k>3$ we can say $k+1>4$.
