m^0 = 1
m^n = m^(n-1).
And now 0^0 = 1 follows immediately, just as it does for 0+0 = 0 and the recursive definition of addition in terms of successor.After that, I consider the question of what x^y means for non naturals as motivated largely by considerations of algebraic and analytic extension. We get negative exponents simply by extending the operation with the requirement that
m^0 = 1
m^(n+p) = m^n*m^p.
We get rational exponents by extending the operation with the requirement that (m^n)^p = m^(n*p).
We get real exponents by extending the operation to the limits where they exist. We get complex exponents by extending the operation to the largest analytic extension.The tension then is that the real and complex extensions don't "confirm" the original reasonable setting of 0^0=1, but I personally find it more objectionable poking a hole in the function, when it was so clear in the case of natural numbers that 0^0 should be 1 (and see the set theoretic and category theoretic accounts for more reasons why this is clear).
`A and B implies A' and so on.
In the end a function, in the mathematical sense, is just a long (possibly infinite) table of `from' and `to' values. We could argue that one definition of ^ is more or less complex than another, for some definition of complexity, but I'm not sure which would win, and if it would be useful.
Your definition of ^ doesn't directly tell me that 2^1=2. It does tell me however that 0^0=1.
exp(x) = 1 + x/1! + x*x/2! + x*x*x/3! + ...
The ^ operator is defined as:
Or similarly set theoretically.