$|\{\text{functions } \{0,1\} \to \mathbb{N} \}|= |\mathbb{N}^{\{0,1\}}|= |\mathbb{N}^2|$
Indeed, if you are thinking of $\mathbb{N}^2$ as the set of ordered pairs of natural numbers, then the function
$\varphi: \{\text{functions } \{0,1\} \to \mathbb{N} \} \to \mathbb{N}^2$
defined as $\varphi(f)= (f(0),f(1))$
is a bijection.
It is an injection: if $f\not=g$, then $f(0)\not=g(0)$ or $f(1)\not=g(1)$, hence $\varphi(f)\not= \varphi(g)$.
It is a surjection: given $(m,n)\in \mathbb{N}^2$, define $f:\{0,1\} \to \mathbb{N}$ as $f(0)=m$, $f(1)=n$. Then $\varphi(f)=(m,n)$.
In fact, for some intuition, think that what this bijection is saying is that a function $\{0,1\}\to \mathbb{N}$ is completely determined by a couple of natural numbers. Who could they be but $f(0)$ and $f(1)$? :)
So, now I claim that $|\mathbb{N}^2|=|\mathbb{N}|$. Indeed, it is the cartesian product of two countable sets, and the cartesian product of a finite number of countable sets is countable, hence the set of functions $\{0,1\} \to \mathbb{N}$ is countable, i.e. it is equinumerous to $\mathbb{N}$.
You comment you can't use cardinal arithmetic: a more constructive proof would be to exhibit a bijection. See this question: How does one get the formula for this bijection from $\mathbb{N}\times\mathbb{N}$ onto $\mathbb{N}$? which attacks the problem of showing that $|\mathbb{N}\times \mathbb{N}|=|\mathbb{N}|$ more directly.
ADDED: An obvious generalization of the above leads to the fact that, for $n\in \mathbb{N}$, $|\{\text{functions } \{0,\dots,n-1\} \to \mathbb{N}\}| = |\mathbb{N}^n|$. Having this in mind, now forget you ever defined $\mathbb{N}^k$ for any $k\in \mathbb{N}$. Let's have a different approach.
Let $I$ be any given set. Define $\mathbb{N}^I$ as $$\mathbb{N}^I := \{\text{functions } I \to \mathbb{N}\}$$
This is the usual notation for the set of functions from $I$ to $\mathbb{N}$.
Now, in set theory (Zermelo-Fraenkel), the number $k\in \mathbb{N}$ is defined as the set $\{0,\dots,k-1\}$. So what if we take $I=\{0,\dots,k-1\}$? We have that $$\mathbb{N}^k=\mathbb{N}^{\\{0,\dots,k-1\\}} = \\{\text{functions } \\{0,\dots,k-1\\} \to \mathbb{N}\\}$$
In particular, with $k=2$, we obtain that $\mathbb{N}^2=\{\text{functions } \{0,1\} \to \mathbb{N} \}= \{\text{functions } 2 \to \mathbb{N}\}$. Now the above proof shows that our new $\mathbb{N}^2$ is equinumerous to the set of ordered pairs of natural numbers. This is nice. Why?
Well, first observe that equinumerous sets are indistinguishable from the point of view of set theory (in fancy words: bijections are isomorphisms in Set). So if two sets are equinumerous, there is no harm done with taking any of them as the definition of the set.
Then, observe that this definition is in much bigger generality. We have defined $\mathbb{N}^I$ for any set $I$. With our previous definition, we could only extend the generality for any finite set $I$, namely defining $\mathbb{N}^I$ as the set of $|I|$-uples of natural numbers: this definition makes no sense a priori for infinite $I$.
A final observation: in this final addendum, $\mathbb{N}$ can be taken to be any set, of course.
That's a good question, but you're not correct.
The reason is that if $f\colon\varnothing\to S$, then $f=\varnothing$. Therefore $\breve f=\varnothing$. Therefore $f=\breve f$ and both satisfy the condition for being a function.
It is true, however, that $\breve f$ is not a function whose domain is $S$ (unless $S$ is empty). And note that we shouldn't require it is a function from $S$, because then the function $f(n)=n+1$ as a function from $\Bbb N$ to itself is not injective anymore, which is plain preposterous.
Also, there is but one unique empty function. The empty set.
Best Answer
Using set theory a function from $X$ to $Y$ is defined as a functional and serial binary relation in $X \times Y$, that is, a subset $R\subset X \times Y$ with the following two properties:
(seriality) For every $x\in X$ there is some $y\in Y$ such that $(x,y)\in R$.
(functionality) If $(x,y)$ and $(x,z)$ belongs to $R$ then $y=z$.
Now observe that if $Y=\emptyset $ then $X\times Y=\emptyset $ and the unique subset of the empty set is the empty set itself so, if any, a function must be represented by the empty set, however in this case the seriality condition will fail. Then we conclude that $Y^X$ is empty.