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The problem: Prove that for $n \in \mathbb N$:

$$ \left(1 + \frac{1}{n} \right)^n = 1 + \sum_{m=1}^{n} \frac{1}{m!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right). $$

The hint is to use the binomial theorem. So the left side can become:

$$ \sum_{m=0}^{n} \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m $$

I don't really know where to go from here, I've tried manipulating the expressions to make them look similar but I'm not really getting anywhere.

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Here's a hint: $$\sum_{m=0}^n \frac{n!}{m!(n-m)!}\left(\frac{1}{n}\right)^m = \sum_{m=0}^n \frac{1}{m!} n(n-1)\dots(n-(m-1))\left(\frac{1}{n}\right)^m$$ –  Dimitrije Kostic Nov 14 '11 at 4:34

2 Answers 2

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Take your second sum $$ \sum_{m=0}^{n} \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m $$ and write it as $$ 1+\sum_{m=1}^n \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m $$ to get the indices to match.

In your first sum $$ \sum_{m=1}^{n} \frac{1}{m!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right) $$ ignoring the $\frac{1}{m!}$ for now, notice $$ \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right)=\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots\left(\frac{n-m+1}{n}\right). $$ Multiplying by $1=\frac{n}{n}$ gives $$ \frac{n}{n}\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots\left(\frac{n-m+1}{n}\right)=\dots $$ Can you take it from there?

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Hint. $\displaystyle 1 - \frac{k}{n} = \frac{n-k}{n}$. So $$\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right) = \frac{(n-1)(n-2)\cdots(n-(m-1))}{(n)(n)\cdots(n)}.$$

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