# How to prove $\frac{m!}{n!} \geq n^{m-n}$

How to prove the following:

$$\frac{m!}{n!} \geq n^{m-n}$$

In my book it's written: "easy to prove by separately considering the cases $$m \geq n$$ and $$m).

I tried using the bounds of Stirling and I got:

$$\frac{m!}{n!} \geq \frac{\sqrt{2\pi}}{e} n^{m-n}$$

But this bound is not tight as the first since $$\frac{\sqrt{2\pi}}{e}\approx 0.92$$

Thanks!

• Closing this question? I don't understand... – Robert Z Jan 25 at 17:47

Stirling approximation is not useful here. The definition of factorial is all we need.

Note that if $$m\geq n$$ then $$m!=\underbrace{m\cdot (m-1)\cdots (n+1)}_{\text{m-n factors each one >n}}\cdot n!$$ On the other hand if $$n>m$$ then $$n!=\underbrace{n\cdot (n-1)\cdots (m+1)}_{\text{n-m factors each one \leq n}}\cdot m!$$ Can you take it from here?

• I upvoted you Robert! Thanks! – Felipe Jan 25 at 17:55
• Thanks. You are welcome! – Robert Z Jan 25 at 17:57

For $$m \ge n$$ $$\frac{m!}{n!}=(n+1) \cdot (n+2) \cdots (m-1) \cdot m \ge n \cdot n \cdots n = n^{m-n}$$

Hint:

For $$m \ge n$$, $$\frac{m!}{n!} = m\times(m-1)\times\cdots\times(n+1) \ge n\times n \times \cdots n=n^{m-n}$$ What happens for $$m \le n$$? $$\frac{m!}{n!} = \frac{1}{n\times(n-1)\times\cdots\times(m+1)} \ge ?$$

No need for Stirling here, elementary computationw work way better.

When $$n \leq m$$, $$m!/n!$$ is a product of $$n-m$$ integers, all greater than $$n$$, thus $$m! \geq n^{m-n}n!$$.

When $$n > m$$, $$m!/n!$$ is a product of reciprocals of $$n-m$$ integers all $$\leq n$$, so the product is $$\geq (1/n)^{n-m}=n^{m-n}$$.