That's actualy neither! I like it though. My proof expaned the darn polynomial via induction, showing that the coefficents were all one. Then, I subtracted that expansion from the geometric series for . The slicker proof is just a multiplication by (1-x) ! You didn't have to multiply by that product. Of course, you can do a lot to the exponent and maintain that inequality!
If we expand the infinite product , a typical term is , where is a sum of powers of two.
The following is the Basis Representation Theorem for base two.
Every positive integer can be represented uniquely as the sum of powers of two. Example: .
Since the exponents of the terms in the infinite series consist of all possible combinations of powers of two, it follows from the theorem that the exponents range over all the positive integers.
We have , i.e. .
So, a third one?