Question: Show that for a set $A$, $|^A 2|$ = $|^{|A| }2|$.
Comments: This is a part of a larger problem that I'm attempting to solve and it seems like the equality is true, but I'm not quite sure how to prove it. Idea: I need to find a bijection from $|^{|A| }2|$ to $|^A 2|$ and I can use that there is a bijection from $|A|$ to $A$, but I'm not sure how to proceed from here. Any input appreciated.
A function $A\to 2$ is a function which, for each element of $A$, decides whether it goes in bin $1$ or in bin $2$. Or whether it is selected or dropped. Therefore it is a way to formalize any subset of $A$, and therefore an element of the power set of $A$. The set of all such functions is therefore the powerset of $A$, often denoted as $2^A$.
A function $A\to\lvert A\rvert$ maps a set to its cardinality. That cardinality is a cardinal number, and as such is considered to be a set of ordinal numbers. The cardinality of $A$ is the cardinality of $\lvert A\rvert$ itself.
For the cardinality of a powerset, it does not matter what the underlying set is. Only its cardinality is important. To illustrate this, consider the task of choosing items from a bag containing $10$ distinct items. There are $2^{10}$ ways to choose items, since each single item may be chosen or not chosen. It does not matter at all what items you are choosing from, be it apples, nuts or whatever.
To formalize things, you'll need three steps:
$$ \bigl\lvert A\to 2\bigr\rvert \quad\leftrightarrow\quad A\to 2 \quad\leftrightarrow\quad \lvert A\rvert\to 2 \quad\leftrightarrow\quad \bigl\lvert\lvert A\rvert\to 2\bigr\rvert $$
The outermost bijections are simply due to the fact that a cardinal number has as many elements as the cardinality it signifies. The inner bijection does the same on the arguments of the domain of the function.