TITLE: No difference between $0/0$ and $0^0$?
QUESTION [5 upvotes]: I have seen discussions about both $0/0$ and $0^0$ and they differ a bit in the way that most seem ok with calling $0/0$ "undefined", while the $0^0$ discussion still seems like a dispute.
If this is correct:
$x/x = x^1 / x^1 = x^{1-1} = x^0$
$x/x = x^0$
$0/0 = 0^0$
Shouldn't they both be equally undefined?

REPLY [2 votes]: 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.