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?

Solution 1:

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 $$\begin{align}k!+3k^2+3k+1 &< k!+(k)k^2+(k^2)k+(k^3)\cdot 1 \\ &= k! + 3k^3 \\ &<k! + k\cdot k^3 \\ &<k! + k\cdot k! \tag{Induction Hypothesis} \\ &= k!(k+1) \\ &= (k+1)! \end{align}$$