sequences and series - Deriving Sum of a Geometric Progression

I was trying to derive the sum of a geometric progression:

$$\sum_{i=0}^{n-1}{ar^i}$$
This is equal to:
$$ar^0+ar^1+ar^2+\cdots+ar^{n-1}$$
Factorising with $a$, I get:
$$a \sum_{i=0}^{n-1}{r^i}$$

Assuming $r$ is $6$ for example, I get:

$$a\left[r^0(1+r(1+r(1+r(1+r(1+r)))))\right]$$
The number of $(1+r)$s inside is equal to $n-1$.

I have no idea how to expand the above equation, and simplify it.

Can someone please give me a step-by-step expansion and simplification of the equation. Something a $9^{th}$ grader would understand.

EDIT
In general, I don't understand how to expand nested braces(recursion isn't the most intuitive thing for me) so a very simple explanation of how this is done will be much appreciated.