Prove that $n! \geq n^3$, for all $n \geq 6$

Question

I want to prove that $n! \geq n^3$, for all $n \geq 6$. I tried with induction, but it doesn't seem to help as it gets even more complicated.

Answer 1

Hint: For $n\geq 6$, $$ n!\ge n(n-1)(n-2)(n-3)\geq 3n(n-1)(n-2) $$ and $$ 3n(n-1)(n-2)-n^3=2n^3-9n^2+6n\geq \ldots>0. $$

Answer 2

Alternate argument (sometimes it's helpful to see these things can be done different ways) -

It's straightforward to verify that for $n \geq 6$, the inequality

$$ (n-1)! \geq 7 \geq 3 + \frac{3}{n} + \frac{1}{n^2}$$

holds. Then rearrange to obtain

$$ n \cdot n! \geq 3n^2 + 3n + 1.$$

After checking for the base case that $k! > k^3$, assume it holds for the case $n$. Now add the two inequalities.