Can you prove why X^0 always = 1?

Sure, piece of cake.

Any number divided by itself is 1. 4/4 = 1, 100/100 = 1, 75.4/75.4 = 1, and so on.

This means that

1 = (x^n) / (x^n), right?

When you divide exponents, you actually subtract them.

So (x^n) / (x^n) is the same as x^(n-n)

x^(n-n) = x^0

Put it all together, and you get this:
1 = (x^n) / (x^n) = x^(n-n) = x^0

1 = x^0

e.g.
x^5/x^5 (x = 2)
= 2^(5 - 5) (= 2^5/2^5)
= 2^0 (= 32/32)
= 1

You can prove it using the multiplication rule

(A^x)*(A^y) = A^(x+y)
(A^x)*(A^0) = A^x

therefore A^0 = 1

Let x^0 = y

Taking log on both the sides

0.log x = log y
=> log y = 0
=> log y = log 1
Therefore y = 1 => x^0 = 1

Take logs of both sides.

log(x^0) = log(1)
0 log(x) = 0 = log(1)

Actually 0^0 is not 1. ~~~ x^1/x^1 = x^0 = 1 is a good proof for non-zero Xs.

Ah, an existance-and-uniqueness proof. I'll state it: The "factoralary" representation of every natural number n exists and is unique. It would be tough to do a whole proof here in plain text, but my gut just tells me to do a proof by contradiction for the uniqueness part. That is, assume that a number n has 2 distinct representations in factoralary form, and then show some kind of contradiction. The existance part seems kind of simple...every time you add 1 to a number, you increase the 1! digit by 1. If that causes overflow, you 'carry' it to the 2!, and make the 1! digit 0. If there's overflow in the 2! place, carry THAT one to the 3! place, making the 2! zero. A formal proof of this kind of smells like it would best be served by induction. I swear I saw a problem like this. Perhaps "The Art of Problem Solving" - volume I or II? It was how I learned all my geeky number theory for high school math competitions back in the day.... EDIT - I found a wikipedia article on this base...it sort of skirts around your question, but it does touch on it. It also adds a '0' to every digit, and has a 0!s place. EDIT of EDIT - DOH. I was beaten to the punch. :(

X^0 = 1 because..

if u reverse the problem..

it will be logx 1 = 0!

there! but if u are still wondering..

maybe the ancient people discovered it...they discovered lots of things that still are a mystery to us..maybe even this!

;)

BY definition, x^n is x multiplied by itself n times

and

x^(-n) is 1 over x^n

According to the rules of of exponential multiplication/division, when you multiply 2 exponential numbers together (when they have the same base), you ADD the exponents. Thus x^m * x ^ n = x^(m + n)

Likewise, when you divide, you subtract the exponents

If you have x^n * x^(-n), this is the same as x^n/x^n which is obviously 1. Therefore, x^n * x^-n must equal 1. But according to the rules of exponential multiplication above, x^n * x^-n =
x^(n-n) = x^0

Therefore x^0 = 1

as a demonstration:

x^5/ x^3 = xxxxx/xxx = xx = x^2 = x^(5 - 3)

x^4 / x^3 = xxxx/xxx = x = x^1 = x^(4 - 3)

x^3 / x^3 = xxx/xxx = 1 = x^0 = x^(3 - 3)
(and continue for negative exponents)

x^2 / x^3 = xx / xxx = 1/x = x^-1 = x^(2 - 3)

looking for patterns is a valuable tool in math!

and as said above, 0^0 is not 1