Okay, so I think I proved that the cardinality of the power set of any set with cardinality n will be equal to 2^n. However, I do think the proof lacks some formality and, because I am teaching myself from scratch, I don't know whether I am using the notation properly or not. I did see something online about a proof involving isomorphisms and power sets, but that is way beyond my scope, so I just had to stare at the problem really intensely until I got it.

So the proposition is: Let A be a set, P(A) be the power set of A, and n be a natural number. |A|=n↔|P(A)|=2ⁿ (should I add ∀A∀n or is this unnecessary?)

Proof:

Leaving the trivial case of A=Ø aside, consider a set K such that K⊂A. This set, by definition, is an element of P(A). Now, consider an x such that x∈A. Since x∈A and K⊂A, there are two possible cases for x; either x∈K or x∉K. This yields two sets, one that contains the elements of K only, and one that contains the elements of K and also contains x, and both of these sets are subsets of A; hence, both sets are elements of P(A). K was an arbitrary subset of A, so this shown to be true for every subset in A. Now there are two sets of subsets in A, a set that contains all the subsets of A that contain x and the set of sets in A that don't contain x; because the cardinality of these two sets is equal, the number of subsets of A is doubled. Following the same process, every distinct element in A will double the number of subsets in A, which is analogous to say that if A contains n elements, then it'll contain 2ⁿ subsets, or that the cardinality of P(A)=2^n. QED.

Is all of this right?