[math-fun] odd digits in n and n^3

Certainly we can prove this theorem for base 2, can't we? How many cubes are there of the form 2^n-1? Something is telling me, "one," but I can't prove it.

--Yes, Catalan's conjecture, nowadays theorem, says A^n - B^m = 1 has only one solution 3^2 - 2^3 = 9-8 = 1 if A,B,n,m, all >=2. Hence 2^n-1 is never a cube if n>=2.