Prove that $\binom{n}{k}$ $n \rightarrow \infty$ is $\frac{n^k}{k!}$ The limit of $\binom{n}{k}$ when $n \rightarrow \infty$ is 
$\frac{n^k}{k!}$, for a fixed $k$.
Intuitively, if n is large, i.e., $n= 80000$ then $\binom{80000}{4} 
=\frac{80000·79999·79998·79997}{k!}$ $\simeq \frac{80000^4}{k!}$
How to formally prove it?
 A: Perhaps you mean to show
$$
\lim_{n\to\infty}\frac{\binom{n}{k}}{\frac{n^k}{k!}}=1
$$
in which case your intuition goes through perfectly:
$$
\frac{\binom{n}{k}}{\frac{n^k}{k!}}=\frac{n(n-1)(n-k+1)}{n^k}=1\Big(1-\frac{1}{n}\Big)\cdots\Big(1-\frac{k-1}{n}\Big)\to 1^k=1
$$
as $n\to\infty$.
A: I am going to assume you are trying to prove an asymptotic.  In that case, take as a hint that for fixed $k$, the numerator is $$n(n-1)\cdots(n-k)=n^k(1-1/n)(1-2/n)\cdots(1-k/n).$$ You can divide both sides by $n^k$, and now apply the squeeze theorem on the right hand side. Note that the fact $k$ is fixed is important.
A: An alternative to yurnero's excellent answer is the following.
Note that the term $x_n:=n!/(n-k)!$ has exactly $k$ many factors, and we can see that $y_n:=n^k$ converges to the same limit by a "catching-up argument", i.e. that that $y_n\geq x_n$ and $x_{n+k}\geq y_n$. We then get that
$$\binom{n}{k}=\frac{n!}{(n-k)!k!}=\frac{x_n}{k!}$$
has the same limit as
$$\frac{y_n}{k!}=\frac{n^k}{k!}$$
when $n\to\infty$.
A: This not (at all) a formal solution.
Considering $$a_n=\binom{n}{k}=\frac{n!}{(n-k)!\,k!}$$ $$\log(a_n)=\log(n!)-\log((n-k)!)-\log(k!)$$ Now, using Stirling approximation for large values of $p$ $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left(p\right)\right)+O\left(\frac{1}{p}\right)$$ Apply this formula to $\log(n!)-\log((n-k)!)$ and use Taylor again for large values of $n$ to get $$\log(n!)-\log((n-k)!)=k \log \left({n}\right)+\frac{k-k^2}{2
   n}+O\left(\frac{1}{n^2}\right)$$ Now, using $a_n=e^{\log(a_n)}$ and Taylor again $$a_n=\frac{n^k}{k!}\left(1+\frac{k-k^2}{2
   n}+O\left(\frac{1}{n^2}\right)\right)$$ 
