Justification of an Algebraic Manipulation

While reading a proof i came across this step which i could not understand. The chunk below is part of bigger expression, but in the interest of reducing noise i am just posting the sub expression if needed please let me know and i'll post the full expression.

$$\left| \dfrac {1} {n\left( c+n-1\right) }\right| =\frac1{n^2}\left| 1-\dfrac {c-1} {n}+O\left( \dfrac {1} {n^{2}}\right) \right|$$

What result is being used to accomplish this ?

Edit: $\left| \dfrac {\left( a+n-1\right) \left( b+n-1\right) } {n\left( c+n-1\right) }\right| = \left| 1+\dfrac {a-1} {n}\right| \left| 1+\dfrac {b-1} {n}\right|\left| 1-\dfrac {c-1} {n}+O\left( \dfrac {1} {n^{2}}\right) \right|$

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You've blocked too much noise. There should be a $1/n$ in factor of the RHS. – D. Thomine Mar 2 '12 at 22:22
thanks i 'll edit it in a sec – Hardy Mar 2 '12 at 22:22
You did edit, but you didn't fix the error. There should still be factor of $1/n$ in the first equation. – joriki Mar 2 '12 at 22:34
You have two contradictory equations, and I was under the impression that the second, correct one is the one given by the authors. There's nothing wrong with that one; I was merely pointing out that the first one is still as wrong as it was before. – joriki Mar 2 '12 at 22:48
I fixed the error since it seemed like you weren't going to do it and it's better not to have wrong equations sitting around. – joriki Mar 6 '12 at 12:36

Divide the top by $n^2$, simplifying as per the right-hand side. Divide the bottom by $n^2$. We want to study the behaviour of $$\frac{1}{1+\frac{c-1}{n}}$$ for $n$ large. Temporarily, let $\frac{c-1}{n}=x$. Note that $\frac{1}{1+x}$ has the familiar power series expansion $$\frac{1}{1+x}= 1-x+x^2-x^3+ x^4-x^5+\cdots.$$
Thus $\dfrac{1}{1+x}=1-x +O(x^2)$, which is exactly what you need.