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I'm looking at one of my professor's calculus slides and in one of his proofs he uses the identity:

$\left(\frac{n+1}{n-1}\right)^n = \left(1+\frac{2}{n-1}\right)^n$

Except I don't see why that's the case. I tried different algebraic tricks and couldn't get it to that form.

What am I missing?

Thanks.

Edit: Thanks to everyone who answered. Is there an "I feel stupid" badge? I really should have seen this a mile a way.

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    $\begingroup$ Because $n+1=(n-1)+2$. $\endgroup$ – vadim123 Feb 23 '15 at 14:39
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Just write $$ \left(\frac{n+1}{n-1}\right)^n = \left(\frac{n-1+2}{n-1}\right)^n =\left(1+\frac{2}{n-1}\right)^n $$

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HINT:

$1+\frac{2}{n-1}=\frac{n-1}{n-1}+\frac{2}{n-1}=\frac{n+1}{n-1}$

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Note that $$ \frac{n+1}{n-1} = \frac{n-1+2}{n-1} = \frac{n-1}{n-1} + \frac{2}{n-1} = 1 + \frac{2}{n-1}. $$

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Hint:

$ (1 + \frac {2}{n-1})^n = (\frac {n-1 +2}{n-1})^n = (\frac{n+1}{n-1} )^n $

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