# Proof or refute $n^n = \Omega(n!)$ with the help of Stirling's approximation

I'm trying to proof/refute the following equation:

$$n^n = \Omega(n!)$$

Generally I would try to use Convergence Criteria and or l'Hôpital's rule to solve such a problem.

$$\lim_{n\to \inf}{{f(n)}\over{g(n)}} = K$$

However, in this case $n!$ is somewhat of a party-stopper. I found Stirling's approximation which states that:

$$n! \approx \left(\frac{n}{e}\right)^n\sqrt{2πn}$$

My idea was therefore:

$${n^n}\over{(\frac{n}{e})^n\sqrt{2πn}}$$ $${n^n}\over{\frac{n^n}{e^n}\sqrt{2πn}}$$ $${e^n}\over{\sqrt{2πn}}$$ $$\lim_{x\to \infty}{{e^2n}\over{2πn}} = \infty$$

$$0 < K \leq \infty$$

therefore the original equation is true.

Is that approach "ok", did I understand Stirling's approximation correctly?

• You don't need Stirling here. You could simply use the definition of $n!$ to prove that $n^n \ge n \cdot n!$. May 11, 2014 at 9:24
• What you actually show is $n^n = \omega(n!)$. May 11, 2014 at 13:30

Stirling's approximation states that $$n! \sim \sqrt{2\pi n} (n/e)^n.$$ This notation means that the ratio between the two sides tends to 1 as $n$ tends to infinity. For your purposes, we can simply write $$n! = \Theta(\sqrt{n} (n/e)^n).$$ Since $\sqrt{n} = o(e^n)$, $$n! = o(n^n).$$
If all you want to show $n! = O(n^n)$, then as Jukka mentions you can use the simple bound $n! \leq n^n$.