Why does $ \frac{b^n-a^n}{b-a}=\sum_{k=1}^nb^{n-k}a^{k-1}$? Trying to work through the answer in this question:
The inequality $b^n - a^n < (b - a)nb^{n-1}$

First consider a sample of the equation to be shown. Consider
\begin{align}
a^{2} + ab + b^{2} &= \frac{(b-a)(b^{2} + ab + a^{2})}{b-a} \\
&= \frac{1}{b-a} \, [ b^{3} -a b^{2} + a b^{2} - a^{2} b + a^{2} b - a^{3} = \frac{b^{3}- a^{3}}{b-a}.
\end{align}
Now apply the same logic to the form
\begin{align}
b^{n-1} + a b^{n-2} + \cdots + a^{n-2} b + a^{n-1}
\end{align}
for which the result is
\begin{align}
\frac{b^{n} - a^{n}}{b-a} = \sum_{k=0}^{n-1} b^{n-k-1} \, a^{k}.
\end{align}
