Why is this true $\frac{1-y^n}{1-y}=(1+y+y^2+...+ y^{n-1})$? I have a heard time seeing why is this true
$\frac{1-y^n}{1-y}=(1+y+y^2+...+ y^{n-1})$
Could you show me some kind of proof, or an identity that would me to find this?

$$\begin{align}
\frac{1-y^n}{1-y}
&=\frac{1\color{red}{-y+y}\color{blue}{-y^2+y^2}-y^3+\cdots+y^{n-2}\color{magenta}{-y^{n-1}+y^{n-1}}-y^n}{1-y}\\
&=\frac{(1\color{red}{-y})+y(\color{red}{1}\color{blue}{-y})+y^2(\color{blue}{1}-y)+\cdots+y^{n-2}(1\color{magenta}{-y})+y^{n-1}(\color{magenta}{1}-y)}{1-y}\\
&=\frac{(1-y)(1+y+y^2+\cdots+y^{n-2}+y^{n-1})}{1-y}\\
&=1+y+y^2+\cdots+y^{n-2}+y^{n-1}
\end{align}$$
