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Proof of a combination identity:$\sum \limits_{j=0}^n{(-1)^j{{n}\choose{j}}\left(1-\frac{j}{n}\right)^n}=\frac{n!}{n^n}$

Prove that product of $n(n-1)(n-2)\dots(2)(1)$ i.e.

$$n! = \sum_{i=0}^{n-1}{n \choose i}(n-i)^n (-1)^i$$

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    You can get your $\TeX$ input formatted by enclosing it in single dollar signs (inline) or double dollar signs (displayed).2012-04-11
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    Hint: the $n^{th}$ forward difference of a polynomial of degree $n$ is $n!$ times its leading coefficient.2012-04-11

2 Answers 2