Yes, you did. It may not have been stated explicitly, but it was implied in a couple places:

1. Binomial theorem. Its statement does not conventionally include any caveats about the x^0 case.

2. D(x^n) = n x^(n-1). When n = 1 and x = 0, you get a 0^0 on the RHS. If 0^0 is taken to be 1, the derivative rule holds. Else, the statement gets messy: D(x^n) = n x^(n-1) if (x,n) != (0,1)