x^0 is intuitive, pretty much so:

    x^0 = x^(y - y) = x^y * x^-y = x^y * 1/x^y = x^y / x^y = 1

x^0 isn't defined to be 1, x^0 IS 1.

x^1 isn't defined to be x, x^1 IS x.

The situation is analoguous to multiplication. a X b is defined as a+a+a+... b times. If you see a problem with x^0 and x^1, you should also see a problem with a X 0 and a X 1.

x^0 is 1 because 1 is multiplication's identity element, just like a X 0 is zero because zero is addition identity element: https://en.wikipedia.org/wiki/Identity_element

> x^0 is intuitive

So you say. If it's planly obvious, why do I get so many results for the Google search "why is x^0 1"?

I understand the multiplicative identity argument, but that's a technical explanation, not an intuitive one.

What's the intuitive reason that multiplying something zero times should equal one? If I don't multiply, then I don't have an answer, and zero's closer to nothing than one is. Why should I use the multiplicative identity if I don't multiply, why would that make sense?

Using the multiplicative identity is yet another choice, not an intrinsic property of numbers. It's a very good choice, and there are a lot of reasons why it's a good choice.

https://en.m.wikipedia.org/wiki/Empty_product

"In mathematics, an empty product, or nullary product, is the result of multiplying no factors. It is by convention equal to the multiplicative identity 1 (assuming there is an identity for the multiplication operation in question), just as the empty sum—the result of adding no numbers—is by convention zero, or the additive identity.[1][2][3]"

> x^0 isn't defined to be 1, x^0 IS 1.

Are you sure? How can you show that? How do you define what an exponent is without defining what x^0 and x^1 are? What does it mean to suggest that x^0 is intrinsically 1? Are you absolutely certain that you just aren't so comfortable with the idea that you can't imagine other possibilities?

https://en.m.wikipedia.org/wiki/Exponentiation

"Formally, powers with positive integer exponents may be defined by the initial condition

b^1 = b

and the recurrence relation

b^(n+1) = b^n * b"

It all follows directly from the very definition of the exponentiation operation itself. You can go work the math yourself, I'm not here to educate illiterate people.

Hahahaha! That's an extra unfortunate choice of insult, it really doesn't look good for you. But I hope it made you feel better. Look, I'm honestly sorry if I offended you along the way, that wasn't my intent. If it was my remark about imagination, it might have been poorly worded on my part, but that wasn't directed at you personally or meant as an insult -- it's actually really difficult for all people to see how certain simple ideas were constructed, when you've known them your whole life. You know what an exponent is so well and so thoroughly, you might not have a strong grasp on how it was defined and developed through history. I don't. Another example: it's very hard to imagine life without 0, but 0 didn't always exist, the symbol 0 was given a definition, and we still haven't fully resolved how to use it in all cases.

Now, since I just quoted the definition of exponentiation itself from Wikipedia and the formal definition includes the base case b^1 = b, I think you've completely failed to make an educated point to go along with your insult. I didn't ask for you to define exponentiation so you could educate me, I asked so that you could think carefully and tell me if you can define what exponentiation is without using the base case. Were you to actually try, you may find it difficult. Or not, I might be wrong, so feel free to prove me wrong or cite a source that proves me wrong, if you want. As it stands, my takeaway for now is that your insult is a substitute for the argument you don't have, so you're forfeiting your position and handing me a walkover.

One problem with this is that

  x^y / x^y 

is restricted to the domain {x | x != 0}. Because e.g. when x = 0 and y = 1, the expression then contains a zero in the denominator. Division by zero is "every number, and therefore no single number" because we can use algebra to "prove" that x / 0 equals "anything we want", almost like the how the principle of explosion works.

If you wanted to prove that 0^0 really does equal 1, you would have to prove that the output of the reduction is unique.