How is $\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$? As the title states, how is: $$\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$$
I can't see the pattern. Can someone please help? Thanks.

Note that for $x\not=1$,
$$(x-1)(x^5+x^4+x^3+x^2+x+1)=x^6-1$$
$$\Rightarrow \frac{x^6-1}{x-1}=x^5+x^4+x^3+x^2+x+1$$
