In MK set theory, I have to show that $B^A = \emptyset$ whenever A is a proper class. A proper class is a class that is not a set i.e. a class that doesn't belong to another class. I thought that I could proove it by contradiction. Suppose $B^A \neq \emptyset$, so by Foundation Axiom there is an $ f \in B^A$ such that $f$ and $B^A$ are disjoint but then i'm stuck. Also I don't understant it intuitively : if i take all the ordered pairs (x,y) for all x in A and a fixed y in B, meaning the function that sends everything in a fixed element in B, could such a class-function belong to $B^A$ ?
EDIT: Just got it! Suppose there is an $f \in B^A$. Then $f$ is a set. $ x \in A$ implise $x \in \{x\} \in (x,y) \in f$ then $x \in (\cup(\cup f)) $. Then A is a subclass of a set, read A belongs to a set (the power set of $\cup \cup f$). a contradiction. Is that a correct proof ?