actually, I thought of a (maybe) helpful way to visualise this.

x^-n is equivalent to 1÷(x^n), so 10^-1 is one tenth, 10^-2 is one hundredth, so on. the number, x, appears in the equation n times.

you can view positive exponents as the inverse, (x^n)÷1. likewise, the number appears n times.

so what happens for x^0? well, zero is neither positive nor negative. and to maintain consistency, x must appear in the equation zero times. so what you're left with is 1÷1, regardless of what number you input as x.

I'd imagine you want something defined recursively like multiplication

• ( 0x = 0 )
• ( xy = x(y-1)+ x ) ( y > 0 ).

So it needs to be

• ( x^0 = c ) (c is some constant)
• ( x^y = xx^{y-1} ) (( y > 0 ) (to see why, replace multiplication with exponentiation and addition with multiplication). So what could ( c ) be? Well, the recursive exponentiation definition we want refers to ( x^0 ) in ( x^1 ). ( x^1 ) must be ( x ) by the thing we wish to capture in the formalism (multiplication repeated a single time). So the proposed formalism has ( x = x^1 = xx^0 = xc ). So ( cx = x ) hence ( c = 1 ), the multiplicative identity. Anything else would leave exponentiation to a zeroth power undefined, require a special case for a zeroth power and make the base definition that of ( x^1 ), or violate the intuition that exponentiation is repeated multiplication.

On an unrelated note, it'd be nice if Lemmy had Mathjax. I just wrote all this on mobile with that assumption, and I'm not rewriting now that I know better.

Indeed, my mind is melted.