Question: Prove that the following identity is valid for all $z$ with $z\neq1$:

$$1+z+z^2+...+z^n=\frac{1-z^{n+1}}{1-z}$$

Answer: 
\begin{align}
S=1+&z+z^2+z^3+\cdots+z^n\\
zS=\quad\quad&z+z^2+z^3+\cdots+z^n+z^{n+1}\\
\\
S-zS=1-z^{n+1}\\
S=\frac{1-z^{n+1}}{1-z}
\end{align}
$$S\_{n+1}=\frac{1-z^{n+1}}{1-z}=\frac{1-z+z-z^{n+1}}{1-z}=1+zS\_{n},$$ with $S\_0=1$.

Then the iterates are

$$S\_0=1,S\_1=1+z,S\_2=1+z+z^2,S\_3=1+z+z^2+z^3,\cdots$$