Rearranging the terms of the series into the usual "descending order" for polynomials, we get a series expansion of:
axn-1 +........ax + a
A basic property of polynomials is that if you divide xn – 1 by x – 1, you'll get:
xn–1 + xn–2 + ... + x3 + x2 + x + 1
That is:
a(xn–1 + xn–2 + ... + x3 + x2 + x + 1) = a(xn-1)/(x-1)
The above derivation can be extended to give the formula for infinite series, but requires tools from calculus. For now, just note that, for | r | < 1, a basic property of exponential functions is that rn must get closer and closer to zero as n gets larger. Very quickly, rn is as close to nothing as makes no difference, and, "at infinity", is ignored. This is, roughly-speaking, why the rn is missing in the infinite-sum formula.