This is ridiculous. Doesn't this seem counter-intuitive for it to be anything else besides 0 or undefined?
Even for things as "objective" as mathematics, definitions ultimately come down to what makes it convenient to manipulate symbols like an expert would.
This comes up all the time in fields as diverse as software engineering, law, finance, business, even hard sciences like physics - things are "right" because they're convenient, and because the consequences of them being that way make it possible to build on those results with new constructs, while doing it a different, more intuitive way would result in those constructs being impossible.
What the article is saying is that your intuition for it being 0 or undefined is because you've been exposed only to exponentiation as repeated multiplication or as the limit of some series; if you consider other theorems like the binomial theorem, and figure out what is necessary for them to hold without special casing, you'll decide otherwise.
Without context, 0 is no more intuitive to me than 1. These two statements are equally intuitive to me, but they give different results for 0^0:
"Zero raised to any power is still just zero."
"Any number raised to the zeroth power is one."
There are 3 cases for 1 and one case for 0 that immediately spring to my mind when considering the problem:
0) Limit of 0^x, as x approaches 0 (from above).
1a) Limit of x^0 as x approaches 0 (from either direction).
1b) Limit of x^x as x approaches 0 (from above).
1c) "What did you multiply by 3 once, to get 3^1? So, multiplying 1 by zero, zero times..."
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.