Proof that $\lim\limits_{h \to \infty} \frac{h!}{h^k(h-k)!}=1 $ for any $ k $ I kind of barely understand this in some way, and I think I would understand it better by a formal proof. Where do I start? 
 A: We have
$$\frac{h!}{(h-k)!}=\underbrace{h(h-1)\cdots(h-k+1)}_{k\;\text{factors}}$$
so
$$\frac{h!}{h^k(h-k)!}=\frac{h(h-1)\cdots(h-k+1)}{h^k}\xrightarrow{h\to\infty}1$$
A: $$\lim_{h\to\infty}\frac{h!}{(h-k)!}\frac1{h^k}$$
$$=\lim_{h\to\infty}\frac{h(h-1)(h-2)\cdots\{h-(k-2)\}\{h-(k-1)\}}{h^k}$$
$$=\lim_{h\to\infty}\prod_{r=0}^{k-1}\frac{h-r}{h}$$
$$=\lim_{h\to\infty}\prod_{r=0}^{k-1}\left(1-\frac rh\right)=1$$
A: Using Stirling's approximation, we have:
$$h! \sim \sqrt{2 \pi h}\left(\dfrac{h}{e}\right)^h$$
Also, for any $k$, we have:
$$(h-k)! \sim \sqrt{2 \pi (h-k)}\left(\dfrac{h-k}{e}\right)^{(h-k)}$$
Therefore: 
$$\dfrac{h!}{h^k(h-k)!}\sim \dfrac{\sqrt{2 \pi h}\left(\dfrac{h}{e}\right)^h}{h^k\sqrt{2 \pi (h-k)}\left(\dfrac{h-k}{e}\right)^{(h-k)}}=A_{k}(h).$$
Simple modifications give us that: $$A_{k}(h)=e^{-k}\sqrt{\dfrac{h}{h-k}}\dfrac{h^{h-k}}{(h-k)^{h-k}}=e^{-k}\sqrt{\dfrac{h}{h-k}}\dfrac{(1-\dfrac{k}{h})^k}{(1-\dfrac{k}{h})^h}$$
Now take the limit of $A_k(h)$ knowing that $\lim\limits_{h\to\infty}((1-\dfrac{k}{h})^h)=e^{-k}$: $$\lim_{h\to\infty}A_k(h)=e^{-k}\sqrt{1}\dfrac{(1-0)^k}{e^{-k}}=1.$$
