# Why $\lim_{n\rightarrow \infty} \frac{n!}{n^{k}(n-k)! } =1$?

I was on brilliant.org learning probability. There was a process explaining how the distribution of a Poisson Random Variable can be obtained from a Binomial Random Variable.

Consider the binomial distribution:

\begin{aligned} P(X=k) &={\binom n k} p^k (1-p)^{n-k}\\ &=\frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} \end{aligned}
Substitute $$m=np$$ , or $$p=\frac{m}{n}$$ : \begin{aligned} P(X=k) &=\frac{n!}{k!(n-k)!} \left(\frac{m}{n}\right)^k \left(1-\frac{m}{n}\right)^{n-k}\\ &=\frac{n!}{k!(n-k)!} \frac{m^k}{n^k} \left(1-\frac{m}{n}\right)^{n}\left(1-\frac{m}{n}\right)^{-k} \end{aligned}
Slightly rearrange \begin{aligned} &=\frac{n!}{n^k(n-k)!} \left(1-\frac{m}{n}\right)^{-k}\frac{m^k}{k!}\left(1-\frac{m}{n}\right)^{n} \end{aligned}

Note that \begin{aligned} & \lim_{n\rightarrow \infty} \frac{n!}{n^{k}(n-k)! } =1,\quad\lim_{n\rightarrow \infty} \left(1-\frac{m}{n}\right)^{-k} =1,\quad \lim_{n\rightarrow \infty} \left(1-\frac{m}{n}\right)^{n} =e^{-m} \end{aligned}

Thus, we have the final result which is equal to the formula for the Poisson distribution.

$$=\frac{m^k e^{-m}}{k!}$$

In all these steps, what I don't understand is the following limit: $$\lim_{n\rightarrow \infty} \frac{n!}{n^{k}(n-k)! } =1$$

• – Martin Sleziak Apr 13 at 7:46
• I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site? – Martin Sleziak Apr 13 at 7:51
• Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them. – billyandriam Apr 13 at 23:13
• billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.) – Martin Sleziak Apr 13 at 23:17

$$\frac{n!}{n^{k}(n-k)! } =\frac{\overbrace{n(n-1)\cdots (n-k+1)}^{k\; factors}}{n^k}= 1\cdot \left(1-\frac{1}{n}\right)\cdots \left(1-\frac{k-1}{n}\right)\stackrel{n \to \infty}{\longrightarrow} 1$$
$$a_n=\frac{n!}{n^{k}(n-k)! }\implies \log(a_n)=\log(n!)-k \log(n)-\log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get $$\log(a_n)=\frac{k(1-k)}{2 n}+O\left(\frac{1}{n^2}\right)$$ Continue with Taylor $$a_n=e^{\log(a_n)}=1+\frac{k(1-k)}{2 n}+O\left(\frac{1}{n^2}\right)$$