Right, we want to define 0^0 in a way that fits in with the usual rules of exponentiation. The problem is that 0^n is always 0 for nonzero n, but a^0 is always

Right, we want to define 0^0 in a way that fits in with the usual rules of exponentiation. The problem is that 0^n is always 0 for nonzero n, but a^0 is always 1 for nonzero a. So 0^0 can't continue both of these patterns by being both 0 and 1. One of the nice reasons to choose 1 over 0 for 0^0 is to make the statement "the cardinality of the set of functions from a size n set to a size m set is m^n". There's exactly one function from the empty set to itself.