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?

$(x-y)\sum_{i=1}^n(x^{i-1}y^{n-i})=\sum_{i=1}^{n}(x^{i}y^{n-i}-x^{i-1}y^{n-i+1})=\sum_{i=1}^nx^{i}y^{n-i}-\sum_{i=1}^{n}x^{i-1}y^{n-i+1}=\sum_{i=1}^nx^{i}y^{n-i}-\sum_{i=0}^{n-1}x^{(i+1)-1}y^{n-(i+1)+1}=\sum_{i=1}^nx^{i}y^{n-i}-\sum_{i=0}^{n-1}x^{i}y^{n-i}=x^n-y^{n}$ 
Substitute appropriately...
