Why does $ \frac{1}{1 - z} = 1 + z + z^2 + z^3 + \ldots$ when $|z| < 1$ 
For $|z| < 1$, $$\frac{1}{1 - z} = 1 + z + z^2 + z^3 + \ldots$$

The fact stated above has appeared in many proofs in my complex analysis textbook, but I have no idea why it is true. Can someone elaborate?
Someone suggested I expand $ (1 - z)^{-1} $, which I tried before posting this question. Perhaps I am wrong, but if $ f(z) = (1- z)^{-1} $ then $ f^{(n)}(z) = n(1 - z)^{-(n + 1)} $ which gives me a Taylor expansion of $ \sum\limits_{n = 0}^{\infty} \frac{(1 - z)^{-(n + 1)}}{(n - 1)!} $ 
I don't see how this helps :(
 A: You can show by simple algebra that
$$
(1-z)(1+z+z^2+\ldots+z^N) = 1-z^{N+1},
$$
or equivalently
$$
1+z+z^2+\ldots+z^N = \frac{1-z^{N+1}}{1-z}.
$$
If $|z|<1$ then $z^{N+1}$ converges to zero as $N$ tends to infinity and you are left with
$$
1+z+z^2+\ldots=\frac{1}{1-z}.
$$
A: Consider the formula for n-th sum of geometric progression
$ S_n = 1 + q + ... + q^n $
$ qS_n =q + q^2 + ... + q^{n+1}$
$ (1-q)S_n = 1 - q^{n+1}$
$ S_n = \frac{1-q^{n+1}}{1-q}$
If |q| < 1, then $ |q|^{n+1} \rightarrow 0 $ and then $q^{n+1} \rightarrow 0$ and $S_n \rightarrow \frac{1}{1-q}$
A: Here is a cool calculation that doesn't assume you already know the sum or remember how to sum a geometric series:
First, prove the sum converges (by any sort of comparison test). Denote it by $S(z)$. Then:
$$S(z) = 1 + z + z^2 + \ldots = 1 + z(1 + z + z^2 + \ldots) = 1 + zS(z)$$
giving the required result.
A: This is a Taylor series. The radius of convergence is a standard calculus exercise.
It is also the geometric series formula. The same proof that works in the finite sum case also works for the infinite sum, when it converges.
