Show that for any natural number K, there is an n large enough so that (2^n) > K.
The Attempt at a Solution
K is a natural number -> the smallest possible K would be 0 (lower bound?) and the smallest 2^n is 1 when n = 0, and the upper bound for both sides are infinite. So if I set n=0 and K =0 I get 2^n > K. I'm not sure if this is the right approach.