No difference between $0/0$ and $0^0$?
Solution 1:
First let me say that we can define $0/0$ and $0^0$ to be whatever we want. The question is: would that definition be useful in general? For $0/0$, the answer seems to be no. For $0^0$, it seems most useful to define $0^0=1$
$0/0$
There are two issues at play here. In the absolute, $0/0$ would b the solution to the equation $$ 0x=0\tag{1} $$ Unfortunately any $x$ works in $(1)$, so that definition fails to give any usefule value.
Another place where people encounter $0/0$ is in connection with limits. For example, if $f$ and $g$ are continuous at $x=a$ and $f(a)=g(a)=0$, we cannot simply plug in the value $a$ to get $$ \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f(a)}{g(a)}=\frac00\tag{2} $$ since that leads us back to the problem with $(1)$. However, in many cases, we can determine $\lim\limits_{x\to a}f(x)/g(x)$. Here are a few examples of limits in the form $0/0$: $$ \begin{align} \lim_{x\to0}\frac xx&=1\\[6pt] \lim_{x\to0}\frac{1-\cos(x)}{x^2}&=\frac12\\[6pt] \lim_{x\to0}\frac{x-\sin(x)}{x^3}&=\frac16 \end{align}\tag{3} $$ As you can see from $(3)$, $\lim\limits_{x\to a}f(x)/g(x)$ totally depends on the choice of $f$ and $g$, and again no clear value arises by considering limits of the form $0/0$.
$0^0$
In common usage, $0^0$ is often encountered in set theory as the number of maps from the empty set to the empty set, or as $x^0$ in combinatorics and polynomials. In all of these cases, $0^0=1$ is the proper definition, since there is $1$ map from the empty set to the empty set, and because $$ \lim_{x\to0}x^0=1\tag{4} $$ Certainly, there are limits of the form $0^0$ which do not equal $1$, for example, $$ \lim_{x\to0}|2x|^{1/\log|x|}=e\tag{5} $$ But they do not occur as often as those mentioned above. Furthermore, since there is a problem raising negative numbers to non-integer powers, even defining $x^y$ in a neighborhood of $(0,0)$ is difficult. This is why we usually consider $x^0$, where the exponent is a fixed integer, when talking about $0^0$.
Solution 2:
They're both equally undefined in the sense that you can have limits with indeterminate forms of both these types that evaluate to any real number you want! However, $0^0$ is sometimes taken to be $1$ in certain contexts: for example, when we write Taylor series (and indeed, power series in general), we often write $$f(x) = \sum_{k = 0}^{\infty} a_k x^k.$$ Since this is centered at $0$, we want it to converge at least there, to $a_0$. However, the first term of $f(0)$ is $a_0 0^0$! We use the convention that $0^0 = 1$ here to simplify notation, because it is in some sense the "right" way to interpret the symbol $0^0$ in this context. There are many other contexts where $0^0$ can be thought of as $1$ in a reasonable way, as there are circumstances where it can be thought of something else in an equally reasonable way. However, as you've discovered, it does not make sense to throw around $0^0$ in everyday life, as an expression of the form $0^0$ can indeed be changed into an expression of the form $0/0$.
Solution 3:
Wikipedia has a good discussion of this issue over here.
Strictly speaking, your point is valid. Another common argument along these lines is that the function $f(x,y)=x^y$ has no limit at $(0,0)$, and therefore should not be defined at this point. However, under certain circumstances, it is useful to "define" $0^0$ to be $1$.
An important related notion that comes up in several area is the idea of the empty product, which is sometimes used to justify the statement $0^0=1$.
Similarly, in the context of Lebesgue integration, one often defines $0\cdot\infty$ to be $0$, even though this quantity is classically undefined.
Solution 4:
$\frac{0}{0}$ and $0^0$ are both undefined. For instance the limit of $\frac{\ln(x+1)}{x}$ as $x$ goes to zero is 1, and for $\frac{\sin^2(x)}{x}$, the limit is zero. In other words, the way you define $\frac{0}{0}$ depends very much on the context and there is no general definition.
The same is true for $0^0$. For instance, $x^x$ as $x$ goes to $0$, tends to $1$. But $(e^x-1)^{x}$ goes to $0$ as $x$ goes to $0$.
It is this multiplicity of possibilities that makes the definition of $\frac{0}{0}$ and $0^0$ impossible.
Solution 5:
I don't think the $\frac{x}{x}=x^0$ argument should be used so freely, since division by zero is illegal but exponentiation of or by zero is not. There are good reasons to define $0^0=1$ in particular contexts -- I'd argue that there are good reasons to do so in most contexts (see below). Furthermore, this definition is entirely compatible with all the power laws: $$(a^m)^n = a^{mn}, \quad a^{m+n}=a^ma^n, \quad (ab)^n=a^nb^n, \quad a^{m-n} = \frac{a^m}{a^n}\ (\text{if}\ a^n \ne 0)$$
Examples of when defining $0^0 = 1$ is useful:
Without allowing $0^0=1$ from time to time, you can't write things like $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \quad \text{for all}\ x \in \mathbb{C}$$ without adding the special case $e^0=1$.
It's compatible with the rule that if $A$ has size $a$ and $B$ has size $b$ then the number of functions $A \to B$ is $b^a$.
It's compatible with the rule that $a^n$ is what you get when you multiply $a$ by $a$ $n$ times: if you multiply anything by itself no times at all then you should get the multiplicative identity, which is $1$.
It's required for the identity $\dfrac{d}{dx}(x^n)=nx^{n-1}$ to hold: the gradient of the line $y=x$ at the origin is $1$!
In the end, the difference between defining $0^0=1$, and not doing so, is that in the former case $0^{\alpha}$ is defined for $\alpha \ge 0$ and in the latter case it's defined for $\alpha > 0$. The difference it makes has measure zero!
I was going to post this as a comment, but it was too long.