Dods said:
Why 0? Is there an underlying theoretical reason to define it like this?
##0## is just a notation. There is an underlying reason to define it like this, but it requires some set theory that you're not familiar with.
Recall that ##B^A## denotes the set of all functions from ##A## to ##B##. This is precisely what ##\mathbb{R}^n## means. It is the set of all functions from ##n## to ##\mathbb{R}##.
But, ##n## is not a set, so how does that make sense? Well, the truth is that natural numbers
are sets. They are defined as follows:
##0## is by definition the empty set. So ##0=\emptyset##.
If ##0,...,n## are defined, then we define ##n+1 = \{0,...,n\}##.
So for example, ##1## is defined as ##1 = \{0\} = \{\emptyset\}## and ##2 = \{0,1\} = \{\emptyset, \{\emptyset\}\}##.
So ##\mathbb{R}^n## is the set of functions from ##\{0,...,n-1\}## to ##\mathbb{R}##.
Let's look in some special cases:
If ##n=1##, then we have ##\mathbb{R}^0 = \{f:\{0\}\rightarrow \mathbb{R}~\vert~\text{f is a function}\}##. But such a function ##f## is uniquely determined by giving ##f(0)##. Indeed, if I say (for example) that ##f(0) = 2##, then my function ##f## is uniquely defined by
f = \{(0,2)\}
(recall that functions are subsets of the cartesian product).
So we often identify the function ##\{(0,x)\}## with the real number ##x##. So this leads us to saying that ##\mathbb{R}^1 = \mathbb{R}##. This equality is of course false (the left hand side contains functions, the right hand side contains real numbers). But it should not be read as an equality, but as an identification.
If ##n=2##, then we have ##\mathbb{R}^2## is the set of functions from ##\{0,1\}## to ##\mathbb{R}##. A typical such function is
f = \{(0,x), (1,y)\}
Again there is a canonical identification of this function ##f## and the ordered pair ##(x,y)##. So we often say that ##\mathbb{R}## is the set of ordered pairs (although it actually are functions).
Now, if ##n=0##, then we have ##\mathbb{R}^0## is the set of functions ##f:\emptyset\rightarrow \mathbb{R}##. What in Earth is a function from the emptyset to ##\mathbb{R}##? Does such a beast exist? Yes, it does. Recall that a function is just a subset of the cartesian product (with some properties that don't matter here). So such a function ##f## would be a subset of ##f\subseteq \emptyset \times \mathbb{R}##. But one can prove that ##\emptyset \times \mathbb{R} = \emptyset##. So a function ##f:\emptyset \rightarrow \mathbb{R}## is just a subset from the empty set. There is only one subset from the empty set, namely the empty set itself. So ##\emptyset## is a function from ##\emptyset## to ##\mathbb{R}## (as crazy as it sounds). So ##\mathbb{R}^0## is then the set of all such functions, so
\mathbb{R}^0 = \{\emptyset\} = \{0\}
