Not really - the explanation from the "mathematician" perspective gives some of the rationale, and it makes perfect sense & in line with intuition.
Defining 0^0 as anything but 1 is weird since x^0 is 1 given the other definitions we have adopted, i.e. x^n = x * x * ... * x (n times) for positive integers, (x^n)^(-1) being defined as the unique number such that x^n * (x^n)^(-1) = 1, (x^n)^(-1) * x^n = 1 for non-zero numbers x, and the simple result that (x^n)^(-1) = (x^(-1))^n from proof by induction & the uniqueness condition of the multiplicative inverse, we then get the natural formula that 1 = x^n * x^(-n) = x^(n - n) = x^0 for non-zero numbers. Defining 0^0 = 0 or undefined doesn't agree with the formula given for all non-zero real numbers, and goes against the limit of x^x as x approaches 0 (as detailed in the article), making x^x a discontinuous function at x = 0 if it is defined as another value.
Defining 0^0 as anything but 1 is weird since x^0 is 1 given the other definitions we have adopted, i.e. x^n = x * x * ... * x (n times) for positive integers, (x^n)^(-1) being defined as the unique number such that x^n * (x^n)^(-1) = 1, (x^n)^(-1) * x^n = 1 for non-zero numbers x, and the simple result that (x^n)^(-1) = (x^(-1))^n from proof by induction & the uniqueness condition of the multiplicative inverse, we then get the natural formula that 1 = x^n * x^(-n) = x^(n - n) = x^0 for non-zero numbers. Defining 0^0 = 0 or undefined doesn't agree with the formula given for all non-zero real numbers, and goes against the limit of x^x as x approaches 0 (as detailed in the article), making x^x a discontinuous function at x = 0 if it is defined as another value.