The proof proceeds by induction.
When n = 1 we have (1)(1-x) = (1-x). Clearly true. (Note also 1 = x0 = xn-1, so we can start our induction as we did, at n=1)
Now suppose it is true for n. We endeavor to show it is true for n+1.
i.e. Suppose n > 1, and it is established that (1+x+x2 +. . .+xn-1)(1−x) = (1-xn) (this is the inductive hypothesis). Then
(1 + x + x2 + ... + xn-1 + xn)(1-x)
= (1 + x + x2 + ... + xn-1)(1-x) + xn(1-x) by the distributive law
= (1-xn) + xn(1-x) where the left summand comes from the inductive hypothesis
= (1 - xn+1) by distributing and cancelling.