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}
Multiply out $(1-z)(1+z+z^2+\dots+z^n)$. 

You get $(1+z+z^2+\dots+z^n)-(z+z^2+z^3+\dots+z^{n+1})=1-z^{n+1}$.