prove that $a^{nb}-1=(a^n-1)((a^n)^{b-1}+...+1)$ Prove that $a^{nb}-1=(a^n-1)\cdot ((a^n)^{b-1}+(a^n)^{b-2}+...+1)$
We can simplify it, like this:
$$(a^{n})^b-1=(a^n-1)\cdot \sum_{i=1}^{b}(a^{n})^{b-i}$$
How can we prove this?

$$
\begin{align}
&(x-1)(x^{n-1}+x^{n-2}+x^{n-3}+\dots+x+1)\\
=&\quad\,x^n+x^{n-1}+x^{n-2}+x^{n-3}+\dots+x\\
&\quad\quad\;\:\,-x^{n-1}-x^{n-2}-x^{n-3}-\dots-x-1\\
=&\quad\,x^n\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;\;-1
\end{align}
$$
