Proof that $\lim\limits_{h \to \infty} \frac{h!}{h^k(h-k)!}=1 $ for any $ k $ [duplicate]
Solution 1:
We have
$$\frac{h!}{(h-k)!}=\underbrace{h(h-1)\cdots(h-k+1)}_{k\;\text{factors}}$$ so $$\frac{h!}{h^k(h-k)!}=\frac{h(h-1)\cdots(h-k+1)}{h^k}\xrightarrow{h\to\infty}1$$
Solution 2:
$$\lim_{h\to\infty}\frac{h!}{(h-k)!}\frac1{h^k}$$
$$=\lim_{h\to\infty}\frac{h(h-1)(h-2)\cdots\{h-(k-2)\}\{h-(k-1)\}}{h^k}$$
$$=\lim_{h\to\infty}\prod_{r=0}^{k-1}\frac{h-r}{h}$$
$$=\lim_{h\to\infty}\prod_{r=0}^{k-1}\left(1-\frac rh\right)=1$$