Solution verfication: Prove that for every cardinality $\alpha, \beta, \gamma$, $\alpha^\gamma \leq \alpha^{\beta+\gamma}$.
So I've been trying to solve this the following way:
Let there be $A,B,C$ sets. Lets assign cardinalities so that $|A|=\alpha, |B|=\beta, |C|=\gamma$, and $B \cap C=\emptyset$. So we need to prove $|A^C| \leq |A^{B \cup C}|$.
We know that $|C| \leq |B \cup C|$, so that means there is a surjective function $h: B \cup C \rightarrow C$. Let there be $M:=\{f| f: B\cup C \rightarrow A \}$ and $S:=\{g|g: C \rightarrow A\}$. We define the function $d: M \rightarrow S$ with $d(f(b))=f(h(b))$.
Lets prove it's surjective: Let there be function $g(b) \in S$. So there exists $x \in B \cup C$ so that $h(x)=b$, and so the origin is $f(h(x))$.
I'm not sure at all of this solution... Thanks in advance for any assistance!
Critique of your proof:
Your $d(f)$ should be a function from $C$ to $A$, when $f$ is a function from $B \cup C$ to $A$. So you need to define $d(f)(b)$ for every $b \in B \cup C$. You define it as $f(h(b))$, where $h(b) \in C$, so $f$ is defined for it. Note that $d(f)(b) = f(h(b))$ is the right use of braces, as $d(f(b))$ would be the value of $d$ on the point $f(b) \in A$, which makes no sense.
The proof of surjectivity should be different. Start with an arbitrary function $g \in S$, so $g: C \rightarrow A$. You should find a function $f$ from $B \cup C$ to $A$, such that $d(f) = g$. So we should have for all $x \in B \cup C$ that $d(f)(x) = g(x)$, where we know (by the previous definition) that $d(f)(x) = f(h(x))$. So we should have, that for all $x \in B \cup C$, $f(h(x)) = g(x)$. But if we are defining $f$ for a point $c \in C$ such that there are two different $p,q \in B \cup C$ such that $h(p) = c = h(q)$, then for $x= p$ we should have $f(c) =f(h(p)) = g(p)$ on the one hand, while also $f(c) = f(h(q)) = g(q)$. While $g$ is pre-given (the function we are trying to find a pre-image of) and there is no guarantee that indeed $g(p) = g(q)$ in this case. So I see no easy way to show your map is a surjection, and it won't be in general.
There is however a much easier way:
Note that $C \subset B \cup C$. So any function from $B \cup C$ to $A$ defines a function from $C$ to $A$ by restricting the function to the subset $C$. This does define a function from $M$ to $S$ and this function is surjective: Assume that $A \neq \emptyset$ (as otherwise $\alpha = 0$ and the result is trivial), and pick any $a_0 \in A$; if $g$ is any function from $C$ to $A$, define $f$ on $B \cup C$ (to $A$) by sending points from $B$ to $a_0$ and sending points $c \in C$ to $g(c)$. Then the restriction of $f$ to $C$ is just $g$ again, trivially, so the restriction map is surjective.
There is no need for $h$ this way.