Finding limit of sequence: $\lim _{n \to \infty} {\frac{n!}{n^k(n-k)!}}=1$ $k$ is nonnegative integer.
I want to show that$$ \lim _{n \to \infty} {\frac{n!}{n^k(n-k)!}}=1$$
My try :
$$ \frac{n!}{n^k(n-k)!} = \frac{n}{n} \frac{n-1}{n} \cdots \frac{n-k+1}{n}$$
I wanted use multiplicative rule

If $a_n \to a$ and $b_n \to b$ , then $a_n b_n \to ab$. 

But It is impossible because 
$$ \frac{n!}{n^k(n-k)!} = \frac{n}{n} \frac{n-1}{n} \cdots \frac{n-k+1}{n}$$
It is infinite product.
I want you to help me.
 A: You actually have a product of a fixed number of terms. It is not about whether the product is finite or infinite, but whether the number of terms is fixed (i.e. not varying with the limiting variable). More precisely, you want to prove:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$

$∀k∈\mathbb{N}\ \left( \lim\limits_{n∈\mathbb{N}→∞} \lfrac{n!}{n^k·(n-k)!} = 1 \right)$.

And this follows from your algebraic decomposition and the following fact for $f : \mathbb{N}×\mathbb{N}→\mathbb{C}$:

$∀k∈\mathbb{N}\ \left( ∀i∈[1..k]\ \left( \text{$\lim\limits_{n∈\mathbb{N}→∞} f(n,i)$ exists} \right) ⇒ \lim\limits_{n∈\mathbb{N}→∞} \prod\limits_{i=1}^k f(n,i) = \prod\limits_{i=1}^k \lim\limits_{n∈\mathbb{N}→∞} f(n,i) \right)$.

This fact is a formal statement of:

The limit of a fixed product of terms is equal to the product of their limits if all those limits exist.

And this fact can be easily proven by induction. It should be clear that this does not apply to things like $\lim\limits_{n∈\mathbb{N}→∞} \prod\limits_{i=1}^n \frac{n+i}{n+i-1} = 2$.
A: Let $\Lambda_{n,k}= {\frac{n!}{n^k(n-k)!}}$. You have proved that $$\Lambda_{n,k}=\prod_{i=0}^{k-1}{\left(1-\frac{i}{n}\right)}=\exp{\bigg(\sum_{i=0}^{k-1}{\log\left(1-\frac{i}{n}\right)}\bigg)}$$
Now we use this :$$\forall x, 0\le x < 1,\quad -x-\frac{x^2}{2}\le \log(1-x)\le-x$$
$$\exp{\big(-\frac{(k-1)k}{n}-\frac{(k-1)k(2k-1)}{12n^2}\big)}= \exp{\bigg(-\sum_{i=0}^{k-1}{\big(\frac{i}{n}+\frac{i^2}{2n^2}}\big)\bigg)}\le   \Gamma_{n,k}\le \exp{\bigg(-\sum_{i=0}^{k-1}{\frac{i}{n}}\bigg)}=  \exp{\bigg(-\frac{(k-1)k}{n}\bigg)}$$
Finally both side tends to $1$.
