$$0^0=|\text{Hom}_{\text{sets}}(\emptyset,\emptyset)|=1$$
I don't know how to use the math text so bear with me.
Supposedly according to this $0^0$ by definition is $1$. Which I'm not sure and would love for someone to explain what it means. I know $\emptyset$ is the empty set, so is it like: by definition $0^0$ is the empty set to the power of empty set has only one possibility which makes by definition $1$, so $0^0$ by definition has to be $1$.
$\operatorname{Hom}_{ \operatorname{Set}}(A,B)$ denotes the set of all functions $A \to B$. (More specifically, this notation means all of the morphisms $A \to B$ in the category of sets, where the morphisms are functions; but if you don't know category theory don't worry about that detail.) A function $f: A \to B$ is defined as a subset $f \subseteq A \times B$ such that for every $a \in A$ there is a unique $b \in B$ such that $(a,b) \in f$. If $A = \emptyset$ and $B = \emptyset$, then there is exactly one such function, namely $f = \emptyset$. $f = \emptyset$ vacuously satisfies the criterion for being a function, and furthermore it is the only function: for if $f \neq \emptyset$ then there is some $(a,b) \in f$ and then $a \in \emptyset$, a contradiction.
All of that is to say that $\operatorname{Hom}_{ \operatorname{Set}}(\emptyset, \emptyset) = \{\emptyset\}$ and so $|\operatorname{Hom}_{ \operatorname{Set}}(\emptyset, \emptyset)| = 1$. So what does that have to do with $0^0$? Well, for nonempty finite sets $A$, $B$ we have $\operatorname{Hom}_{ \operatorname{Set}}(A, B) = |B|^{|A|}$. For example there are $2^3 = 8$ functions from $\{1,2,3\}$ to $\{1,2\}$ (we have two choices of where to send $1$, two choices of where to send $2$, two choices of where to send $3$.) So, if we define $0^0 = 1$ then this pattern holds for all empty sets as well. So it's an argument that we ought to define $0^0 = 1$ instead of some other choice. Whether that's a proof or not depends on your definition of $a^b$. If you define $a^b$ so that $a^b$ is the cardinality of $\operatorname{Hom}_{ \operatorname{Set}}(A,B)) = \{\emptyset\}$, where $A = \{1,2, \ldots, a\}$ and $B = \{1, 2, \ldots, b\}$ then $|\operatorname{Hom}_{ \operatorname{Set}}(\emptyset, \emptyset)| = 1$ proves that $0^0 = 1$. If you define $a^b$ in some other way, then this is not a proof, but it nonetheless a convenient fact.
Note that sometimes we denote $\operatorname{Hom}_{ \operatorname{Set}}(A,B)$ by $B^A$, so that (if $0^0 = 1$) we have $|B^A| = |B|^{|A|}$ for $A,B$ are finite sets. I've avoided using the notation $B^A$ above to avoid any confusion with the operation on natural numbers $b^a$.
There are a few tricky things going on here. Rather than landing right on the empty set with a bang, let's try to get there bit by bit.
Suppose we have a set of people $A$ and a set of teams $B$. We'll start with $A = \{\text{Pat}, \text{Quinn}\}$ and $B = \{\text{Rams}, \text{Seahawks}, \text{Tigers}\}$. Each person has a favorite team, and for some reason, we want to know how many different ways the people in $A$ can choose a favorite team from $B$. (They don't have to pick different teams; two people can have the same favorite team.)
One way to answer this question is to just list them:
$$ \text{Pat} \to \text{Rams}, \text{Quinn} \to \text{Rams} \\ \text{Pat} \to \text{Rams}, \text{Quinn} \to \text{Seahawks} \\ \text{Pat} \to \text{Rams}, \text{Quinn} \to \text{Tigers} \\ \text{Pat} \to \text{Seahawks}, \text{Quinn} \to \text{Rams} \\ \text{Pat} \to \text{Seahawks}, \text{Quinn} \to \text{Seahawks} \\ \text{Pat} \to \text{Seahawks}, \text{Quinn} \to \text{Tigers} \\ \text{Pat} \to \text{Tigers}, \text{Quinn} \to \text{Rams} \\ \text{Pat} \to \text{Tigers}, \text{Quinn} \to \text{Seahawks} \\ \text{Pat} \to \text{Tigers}, \text{Quinn} \to \text{Tigers} $$
However, we can be a bit more systematic about this. We can recognize that Pat can choose any of $3$ teams and so can Quinn, independently of Pat's choice. That leads to a total of $3 \times 3 = 3^2 = 9$ combined choices.
This kind of reasoning allows us to answer the question for larger sets $A$ and $B$. Notice that for the purposes of this question, we don't care what the elements of $A$ and $B$ are, only their sizes or cardinalities. We denote these cardinalities with $|A|$ and $|B|$.
For this example, $|A| = 2$ and $|B| = 3$. But suppose we had $|A| = 5$ people and $|B| = 7$ teams. Then each of those $5$ people would have $7$ choices, and the total number of combined choices would be $7 \times 7 \times 7 \times 7 \times 7 = 7^5 = 16\,807$. Just imagine trying to list all those out!
More generally then, the number $N$ of combined choices would be
$$ N = |B|^{|A|} $$
Now, let's start restricting things. If there's only $|A| = 1$ person, then the number of combined choices is just the number of that one person's choices, which is $|B|$:
$$ N = |B|^1 = |B| $$
On the other hand, if there's only $|B| = 1$ choice, then each of the $|A|$ people can only make that one choice, so the number of combined choices is also only $1$:
$$ N = 1^{|A|} = 1 $$
What if there's $|B| = 0$ choices—that is, $B$ is the empty set? Then there are no ways for each person in $A$ to choose a favorite team, so the number of combined choices is $0$:
$$ N = 0^{|A|} = 0 $$
We seem to be on a collision course with $N = 1$ when $|A| = |B| = 0$, which doesn't fit this rule! But there's a twist.
The twist is that if there are no people—that is, if $A$ is the empty set and $|A| = 0$—the number of combined choices is $1$, not $0$. How can this be? There are no people to make any choices; how can there be any combined choices?
The reason there's actually one combined choice is that a combined choice is simply a way to map each person in $A$ to a choice in $B$. When $B$ was the empty set, there was no way to do that mapping because each person had no choice available. But when $A$ is the empty set, there is a way: The empty mapping qualifies, because each person in $A$ is making a choice. Or equivalently, there is no person in $A$ who does not make a choice, because there is no person in $A$ to begin with! Hence the number of combined choices is $1$, not $0$.
This is a rather non-intuitive way to interpret these words in ordinary English, because there's no actual choosing going on. But there is a choice function—namely, the empty function. And it makes actual numerical sense:
In our original example, $2$ people each made one of $3$ choices, so there were $3^2 = 9$ combined choices.
If we take away one of the people (leaving one), then the number of combined choices should be reduced by a factor of $3$, and sure enough, there are $3^1 = 3$ combined choices.
If we take away another person (leaving none), then again the number of combined choices should be reduced by a factor of $3$, and true to form, there is $3^0 = 1$ combined choice.
We can't quite reason this way when $|B| = 0$, because it leads to the expression $0/0$. But recognizing that an empty set $A$ of people can be mapped to any set $B$ in a single way motivates the common assignment $0^0 = 1$.
I've worded this in rather informal language, but it can be formalized and carry forward essentially the intuition given above (hopefully).
