# How can I prove this equation: $(1+ \frac{1}{n})^n = 1+ \sum\limits_{k=1}^n{\frac{1}{k!}}1(1-\frac{1}{n})\cdot(1-\frac{2}{n})…(1-\frac{k-1}{n})$ [duplicate]

Possible Duplicate:
Proving sequence equality using the binomial theorem

$(1+ \frac{1}{n})^n = 1+ \sum\limits_{k=1}^n{\frac{1}{k!}}\cdot1\cdot(1-\frac{1}{n})\cdot(1-\frac{2}{n})\cdot…\cdot(1-\frac{k-1}{n})$

This is how far I came using the binomial theorem:

$(1+ \frac{1}{n})^n = 1 + \sum\limits_{k=1}^n{\frac{1}{k!}\cdot\frac{n!}{n^k\cdot(n-k)!}}$

I don't know how to rearrange that tail to be the same as in the original equation.

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## marked as duplicate by Arturo Magidin, Mike Spivey, Sasha, t.b., Qiaochu YuanNov 14 '11 at 22:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

$$\frac{n!}{n^k(n-k)!}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{n\,n\,n\cdots n}=\left(1-\frac0n\right)\,\left(1-\frac1n\right)\,\left(1-\frac2n\right)\cdots\left(1-\frac{k-1}n\right).$$

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Oh thanks! :) I was so close. – mcb Nov 14 '11 at 20:41
I agree.   – Did Nov 14 '11 at 20:44