Why does $\lim_{n\to\infty} \frac{n!}{(n-k)!n^k}$ equal 1 [duplicate]

$$\frac{(n!)}{(n-k)!n^k}=\frac{n(n-1)(n-2)\cdots\{n-(k-1)\}}{n^k}=\prod_{0\le r\le k-1}\left(1-\frac rn\right)$$

A (non-rigorous) probabilistic interpretation: There are balls of $n$ different colors, of which we choose $k$. The probability that we choose $k$ different colored balls is $$\frac{n(n-1)\dots(n-(k-1))}{n^k} = \frac{n!}{(n-k)!n^k}.$$ It is intuitively clear that as the number of colors increases, this probability goes to $1$.

I myself just found a better method:

$\frac{N!}{(N-k)!N^k}=\frac{\prod\limits_{i=0}^{k-1} (N-i)}{N^k}$

The top therefore is a polynomial of order k with the highest order having coefficient 1. Therefore all other terms vanish and you are left with 1 as N approaches infinity,

Thanks!