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?

For $\ x = a^b\ $ it is $\ f_n \,:=\, \dfrac{x^n-1}{x-1}\, =\, x^{n-1}+ x^{n-2} +\cdots  + x + 1.\,$ Here is a proof by telescopy.
Notice that $\ \color{#0a0}{f_{n} - f_{n-1} = x^{n-1}}\ $ since $\ \dfrac{x^n-1}{x-1} - \dfrac{x^{n-1}-1}{x-1}\, =\, \dfrac{x^n-x^{n-1}}{x-1\quad }\, =\, x^{n-1}$
$\begin{eqnarray}{\rm\! Hence}\ \  &&\qquad\quad\ \ \, \underbrace{\color{#0a0}{f_n}} \\
\,&=&\ \ \overbrace{\color{#0a0}{x^{n-1}}\ \ \ \ \ \ +\ \ \ \ \ \  \underbrace{\color{#0a0}{f_{n-1}}}}\\
\,&=&\ \ x^{n-1} +\  \overbrace{{x^{n-2}\ \ \ \  \ \,+\,\ \ \ \ \ \underbrace{f_{n-2}}}}\\
\,&=&\ \ x^{n-1} +\ x^{n-2}+\ \overbrace{x^{n-3}\ \ \ \ +\ \ \ \ {f_{n-3}}}\\
\,&&\qquad\qquad\ \ \vdots\qquad\qquad\qquad\qquad\qquad\ddots
\end{eqnarray}$
Alternatively we can write it in telescopic form
$\ \ \begin{eqnarray}
\color{#c00}{f_n-f_k}\, =\, 
\underbrace{\phantom{f_n-f_{n-1}}}_{\Large x^{n-1}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\color{#c00}{f_n}}&&\!\!\overbrace{{-f_{n-1}} 
+\underbrace{\phantom{f_{n-1} - f_{n-2}}}_{\Large x^{n-2}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!f_{n-1}}^{=\, 0}&&\!\!\overbrace{-f_{n-2} + f_{n-2} }^{=\, 0}
- \,\cdots\, \overbrace{-f_{k+2} 
+ \underbrace{\phantom{f_{k+2} - f_{k+1}}}_{\Large x^{k+1}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!f_{k+2}}^{=\, 0}&&\!\!\overbrace{-f_{k+1}
+\underbrace{\phantom{f_{k+1} - f_{k}}}_{\Large x^{k}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!f_{k+1}}^{=\,0}&&\!\!\!\color{#c00}{-\,f_k}\\
\end{eqnarray}$
