Why is $\lim_{n\to\infty}\binom{n}{i} \frac{1}{n^i} \left (1-\frac{1}{n}\right)^{n-i}=\frac{1}{i!} e^{-1}$? In my probability script the following equation is used, to show that for $n\to \infty, \displaystyle\binom{n}{\frac{1}{n}}$ converges against the Poisson distribution $Po(1)$. 
$$\lim_{n\to\infty} f_X(i) = \lim_{n\to\infty} \frac{n(n-1)...(n-i+1)}{i!}\frac{1}{n^i}\left(1-\frac{1}{n}\right)^{n-i} = \frac{1}{i!}e^{-1}$$
It is clear to me why for a set $i$,  $\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n-i} = e^{-1}$. However, I am not sure why the two first fractions result in $\frac{1}{i!}$. I thought that the $n(n-1)...(n-i+1)$ and $n!$ cancel out, but that wouldn't be an equality, at least as far as I know. It could also be that my assumption that $\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n-i} = e^{-1}$ in this case.
 A: Well, for $i$ fixed, there are $i$ terms in the product $n(n-1)...(n-i+1)$ and so $\dfrac{n(n-1)...(n-i+1)}{n^i}=\prod_{j=0}^{i-1}(\dfrac{n-j}{n})=\prod_{j=0}^{i-1}(1-\dfrac{j}{n})$.
Now $j/n\to0$ as $n\to\infty$ for each $0\leq j\leq i-1$ so $\prod_{j=0}^{i-1}(1-\dfrac{j}{n})\to 1$ and you get the result.
A: $$a_n=\binom{n}{i} \frac{1}{n^i} \left (1-\frac{1}{n}\right)^{n-i}$$
$$\log(a_n)=\log (n!)-\log ((n-i)!)-\log (i!)-i\log(n)+(n-i)\log\left (1-\frac{1}{n}\right)$$ Now, for large $n$, use Stirling approximation and Taylor series to get (I skip the simplifications)
$$\log(a_n)=\log \left(\frac{1}{e\, i! }\right)-\frac{i^2-3 i+1}{2 n}+O\left(\frac{1}{n^2}\right)$$
Now, use $a_n=e^{\log(a_n)}$
A: $$L=\lim_{n\to \infty} {n \choose k}\frac{1}{n^k}\left(1-\frac{1}{n}\right)^{n-k}$$
$$L=\lim_{n \to \infty} F(n)= \lim_{n \infty} \frac{n(n-1)(n-2)(n-3)...(n-k+1)}{n^k~k!} \lim_{n \to \infty}\left(1-\frac{1}{n}\right)^{-k} \lim_{n \to \infty}\left(1-\frac{1}{n}\right)^{n}$$
$$L= \lim_{n \to \infty} \frac{ (1-1/n)(1-2/n)(1-3/n)...(1-(k-1)/n}{k!}. 1.~ e^{-1}$$
$$\implies L=\frac{e^{-1}}{k!}$$
