Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

marked as duplicate by Arturo Magidin, Mike Spivey, Sasha, t.b., Qiaochu Yuan Nov 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 1

up vote 4 down vote accepted

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

share|improve this answer
Oh thanks! :) I was so close. –  mcb Nov 14 '11 at 20:41
I agree. $ $ $ $ –  Did Nov 14 '11 at 20:44

Not the answer you're looking for? Browse other questions tagged or ask your own question.