solve the equation $1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$ im trying to solve the equation  $$(E)\quad 1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$$
attempt : because $1$ isnt a solution we have  $$\begin{aligned}
1+2 z+2 z^{2}+\cdots+2 z^{n-1}+z^{n} &=2\left(z^{0}+z+z^{2}+\cdots+z^{n-1}\right)-1+z^{n} \\
&=2 \frac{1-z^{n}}{1-z}-1+z^{n} \\
&=\frac{2-2 z^{n}+z^{n}-z^{n+1}}{1-z}-1 \\
&=\frac{2-z^{n}-z^{n+1}}{1-z}-1
\end{aligned}$$ then (E) becomes $$z^{n+1}+z^{n}-z-1=0$$ but i dont know how to get any further.

Hint: \begin{align}
z^{n+1}+z^n-z-1&=z^n(z+1)-(z+1)\\
&=(z^n-1)(z+1)
\end{align}
Also, I think you should justify why the second equality holds, i.e. explain why you can assume $1-z\neq 0$.
