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

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?

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 $m \le n$? $$\frac{m!}{n!} = \frac{1}{n\times(n-1)\times\cdots\times(m+1)} \ge ?$$