Show that $\Sigma_{j=0}^n z^j=\frac{1-z^{n+1}}{1-z}$

Question

As in the question I have to show that $$\sum_{j=0}^n z^j=\frac{1-z^{n+1}}{1-z}$$ So if we suppose that the above is true then clearly \begin{align}(1-z)(1+z+z^2+\dots+z^n)&=(1-z)+(z-z^2)+\dots+(z^{n-1}-z^n)+(z^n-z^{n+1})\\&=1-z^{n+1}\end{align} as required. But what if I didn't know the right hand side of this equation? How would one work it out just from the sum on the LHS? Thanks for the help.

Answer 1

Hint: Set $S_n=\sum_{j=0}^n z^j$ and show that $$S_n−zS_n=z^0−z^{n+1}$$