In proving the equivalence of the series and $\lim_{n\to\infty}(1+\frac{x}{n})^n$ definitions of $e^x$, it turns out that I need to show that for any positive integer $k$, $$\lim_{n\to\infty}\frac{n!}{(n-k)!~n^k}=1.$$
I have no idea how to show this (I'm a Physics student). Any help?
We have that
$$\frac{n!}{(n-k)!~n^k}=\frac{n(n-1)\ldots(n-k+1)}{n^k}=\prod_{i=0}^{k-1}\left(1-\frac in\right)$$
$n! = n(n-1)(n-2)\cdots(n-(k-1))(n-k)!$
So,
$\dfrac{n!}{(n-k)!n^k} =\dfrac{n(n-1)(n-2)\cdots(n-(k-1))}{n^k} = \dfrac{1(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k-1}{n})}{1}$
Now, apply the limit.
Recognize that
$$\lim_{n \rightarrow \infty} \frac{n!}{(n-k)! n^k} = \lim_{n \rightarrow \infty} \frac{n(n-1) \cdots (n-k+1)}{n^k} = \lim_{n \rightarrow \infty} \frac{n}{n} \cdot \frac{n-1}{n} \cdots \frac{n-k+1}{n} = 1,$$
where you first use cancellation of the factorials, then realize that you have the same number of factors in the numerator and denomintor, leading to the expression above, where each fraction approaches 1.
