$\frac{1}{1-x}$ series expansion

How do I know that the expression:

$$\frac{1}{1-x}$$

Is equal to the infinite sum:

$$-\left(\frac{1}{x}\right)-\left(\frac{1}{x}\right)^2-\left(\frac{1}{x}\right)^3-\left(\frac{1}{x}\right)^4+...$$

Thanks!

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\begin{align} \frac{1}{1-x} = & \frac{1}{x\left[\frac{1}{x}-1\right]} \\ = & -\left(\frac{1}{x}\right)\frac{1}{1-\left(\frac{1}{x}\right)} \\ = & -\left(\frac{1}{x}\right)\left[ 1+\left(\frac{1}{x}\right)^1+\left(\frac{1}{x}\right)^2+\left(\frac{1}{x}\right)^3+\left(\frac{1}{x}\right)^4+\cdots\right] \\ \end{align}

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If $|x^{-1}|<1$ then the sum of geometric series: $$-\sum_{n=1}^\infty x^{-n}=-\frac{1}{x}\frac{1}{1-x^{-1}}=\frac{1}{1-x}$$

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Nice work, dear friend! –  amWhy Aug 14 at 11:46

Hint: If I were you, I would try to divide $1$ by $1-x$. This is an elementary way to find those terms.

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Good suggestion, Babak! –  amWhy Apr 13 '13 at 19:08