Dividing by Zero Using Exponents I am sorry if I am missing some fundamental rule disproving my question, but I am very confused here. 
So $0^0 = 1$
and
$x/x = x^0 = 1$
so if $x = 0$
$0/0 = 1$
The problem here is people have told me 0 divided by 0 equals undefined, but by using exponents I have proved this wrong. I have to be doing something wrong here. So, what is wrong with what I just did?
 A: Your reasoning is wrong beginning with the line "so if $x=0$", because division by $0$ is undefined. It doesn't matters whether you use a variable or a composite expression as the denominator, as long as it equals $0$, the argument is incorrect.
As is almost always the case with this type of questions, you seem to ignore the fact that mathematics is a deductive science. We begin from axioms and definitions, then demonstrate that some results logically follow from them. The properties of mathematical objects are not known due to observation of the physical world and are not laws of nature, but consequence of their definitions and the logical framework in which they are made (search for "construction of the real numbers".
It makes sense, however, to ask why for all $x \in R$, $x/0$ is undefined. The answer is in short, because it can't be defined meaningfully (in a way that doesn't contradicts the axioms of the real numbers).
Also, it is incorrect that "0 divided by 0 equals undefined". In ordinary treatments of mathematics, it doesn't equals anything. "is" has more than one meaning, not just the same as "equals". However, in formal treatments of mathematics like Metamath (link is to relevant theorem) the value of a function outside its domain (that is it, where it is not defined) equals an arbitrary object.
A: 0^0 is not equal to 1. It is not defined. 
however:
(Tending to 0+)^(tending to 0) is 1. This is a standard limit. Thus what you are using is wrong.
A: The "fundamental rule" is that you are not allowed to use undefined expressions in your proofs. Using an expression before its definition is a syntactic error, period. If you start your argument with a statement like "$0^0 = 1$" without defining what "$0^0$" is supposed to mean, then it's the same type of error as saying "Let $x = ([[3,,7\bigoplus{\frac{\emptyset}{:)}}$" without specifying what "$([[$" and "$\frac{\emptyset}{:)}$" and all the other symbols are supposed to mean. It's just a meaningless combination of weird characters.
Some of the answers here are also mostly meaningless combinations of characters. Let's take a look at some of them.
The statement "(Tending to 0+)^(tending to 0) is 1", except being an informal combination of English words and mathematical symbols, is also misleading, because it suggests a wrong mathematical statement. "(Tending to 0+)^(tending to 0)" can be whatever you want. For example, take a look at the following limits:
$\qquad \lim_{n\to\infty}\frac{1}{n}^{\frac{1}{n}} = 1$
$\qquad \lim_{n\to\infty}\frac{1}{n}^{\frac{1}{log(log(n))}} = 0$
$\qquad \lim_{n\to\infty}(\exp(-n \log(\pi))^{-\frac{1}{n}} = \pi$
$\qquad \lim_{n\to\infty}\frac{1}{n}^{(\frac{n \mod 2}{n} + \frac{(n+1) \mod 2}{\log(\log(n))})}$ is undefined, the sequence does not converge to anything.
The statement "$0^0\neq 1$" is invalid for the same reasons: it contains an undefined expression "$0^0$", so the whole statement is again just a meaningless sequence of characters.
The comment of Renato Faraone is correct, because there is an explicit limit $\lim\limits_{x\to 0+}x^x$ given.
However, sometimes (for example if one is working with polynomials), one defines $x^0 \equiv 1$, because it makes sense in the context of polynomials, and helps to keep the notation simple yet consistent. However, as soon as you leave the context of e.g. polynomials, you must either abandon this notation, or check that it still makes sense in the other context. For example, $x/x$ is not a polynomial, but rather a rational function. To be precise, "$x/x$" does not have any meaning either, until you specify what it means. For example, you could define a function $x \mapsto \frac{x}{x}$ with the domain $\mathbb{R}\backslash\{0\}$ and the codomain $\mathbb{R}$. Having this definition, we can clearly state that this thing is not a polynomial on $\mathbb{R}$, because it is not even defined at $0$. Therefore, if your previous definition $0^0\equiv 1$ was designed for working with polynomials, you are not allowed to throw it at your weird non-polynomial function, at least not without extending the definition of $0^0$ in some sensible way.
