Simplify the infinite product (1+x)(1+x²)(1+x⁴)(1+x⁸)(1+x¹⁶)..., given |x| < 1.

Answer: 1/(1-x)

Explanation:

Let P = (1+x)(1+x²)(1+x⁴)(1+x⁸)(1+x¹⁶)...

Then (1-x)P = (1-x)(1+x)(1+x²)(1+x⁴)(1+x⁸)(1+x¹⁶)...
   = (1-x²)(1+x²)(1+x⁴)(1+x⁸)(1+x¹⁶)...
   = (1-x⁴)(1+x⁴)(1+x⁸)(1+x¹⁶)...
   = (1-x⁸)(1+x⁸)(1+x¹⁶)...
   = ...
   = 1

Since (1-x)P = 1, it follows that P = 1/(1-x)

Now do you see why I called this problem "Cotton Candy"?  It's the way that factor, (1-x) "eats" through the rest of the factors, which melt as soon as they're touched!

As the correct responses have come in, I've noticed that there are two other ways people approach this problem, which are also quite interesting...

1. Note that the product through (1+x^{(2^(n-1))}) = (1-x^{(2^n)})/(1-x), then take the limit as n-->infinity.

2. Note that the product through (1+x^{(2^n)}) is equal to the sum 1+x+x²+...+x^{(2^n)}, and then when n goes to infinity, the sum equals 1/(1-x).