If your argument is really all that long and involved, it’s probably unnecessarily complicated. We have
$$\varphi:\left(A^B\right)^C\to A^{B\times C}:f\mapsto\varphi_f\;,$$
where $$\varphi_f\big(\langle b,c\rangle\big)=f(c)(b)\;.$$
Suppose that $f,g\in\left(A^B\right)^C$ and $f\ne g$; then there is some $c\in C$ such that $f(c)\ne g(c)$. Now $f(c)$ and $g(c)$ are functions from $B$ to $A$, so there must be some $b\in B$ such that $f(c)(b)\ne g(c)(b)$. But then $\varphi_f\big(\langle b,c\rangle\big)\ne\varphi_g\big(\langle b,c\rangle\big)$, so $\varphi_f\ne\varphi_g$, and it follows that $\varphi$ is injective.
Now let $F\in A^{B\times C}$ be arbitrary. For each $c\in C$ let $F_c=\left\{\left\langle b,F\big(\langle b,c\rangle\big)\right\rangle:b\in B\right\}\in A^B$, and let
$$f:C\to A^B:c\mapsto F_c\;;$$
it’s entirely straightforward to verify that $\varphi_f=F$ and hence that $\varphi$ is also surjective.
By the way, it’s not just that these two sets have the same cardinality that’s important: this specific bijection is important in its own right.
The natural bijection from $A^B\times A^C$ to $A^{B\cup C}$ is defined as follows:
$$\varphi:A^B\times A^C\to A^{B\cup C}:\langle f,g\rangle\mapsto\varphi_{f,g}\;,$$
where $$\varphi_{f,g}:B\cup C\to A:x\mapsto\begin{cases}
f(x),&\text{if }x\in B\\
g(x),&\text{if }x\in C\;.
\end{cases}$$
Note: It’s important to assume here that $B\cap C=\varnothing$.
The details will be quite different, but if you mimic the pattern of the argument that I gave above, you should be able to prove reasonably efficiently that this $\varphi$ is a bijection.
