If $|B|=\mu$ and $|A_i|=\kappa_i \;\forall i\! \in\! I$, than $(\prod_{i\in I}\kappa_i)^\mu =|\text{Fun}(B,\prod_{i \in I}A_i)|$. Also, $ \prod_{i\in I}\kappa_i^\mu = |\prod_{i \in I}{|\text{Fun}(B,A_i)|}| $. I'm not sure this last expression is right, and, above all, I don't know how to make a bijection between the two.
This is my first question, any suggestions are appreciated. :)
The proof of this is essentially the same as the proof that there is a bijection $$\mathrm{Fun}(A, \mathrm{Fun}(B,C)) \to \mathrm{Fun}(B, \mathrm{Fun}(A,C))$$ Each function $f : A \to \mathrm{Fun}(B,C)$ induces a function $\widetilde{f} : B \to \mathrm{Fun}(A,C)$ defined by $(\widetilde{f}(b))(a) = (f(a))(b)$ for all $a \in A$ and $b \in B$, and this establishes a bijection.
[See also: currying—this is the result of uncurrying, swapping the order of the arguments, and then currying.]
Now here it's a bit more complicated, but the idea is the same.
An element of $\prod_{i \in I} A_i$ is a function $g : I \to \bigcup_{i \in I} A_i$ with $g(i) \in A_i$ for each $i \in I$. Thus a function $f : B \to \prod_{i \in I} A_i$ assigns to each $b \in B$ a function $f(b) : I \to \bigcup_{i \in I} A_i$.
We can then obtain a function $\widetilde{f} : I \to \bigcup_{i \in I} \mathrm{Fun}(B,A_i)$ defied by $(\widetilde{f}(i))(b) = (f(b))(i)$ for all $i \in I$ and $b \in B$; note that we really do have $\widetilde{f} \in \prod_{i \in I} \mathrm{Fun}(B,A_i)$.
The assignment $f \mapsto \widetilde{f}$ induces a bijection $$\mathrm{Fun}\left(B, \prod_{i \in I} A_i\right) \to \prod_{i \in I} \mathrm{Fun}(B,A_i)$$
You now need to check the details.