The cardinality of $B^A$ is the same if $A$ [resp. $B$] is replaced with a set containing the same number of elements as $A$ [resp. $B$].
Set $b = |B$|. When $b \lt 2$ there is little that needs to be addressed, so we assume $b \ge 2$. Assume $|A| = n$.
A well known result of elementary number theory states that if $a$ is a natural number and $0 \le a \lt b^n$ then it has one and only one base-$\text{b}$ representation,
$$\tag 1 a = \sum_{k=0}^{n-1} x_k\, b^k \text{ with } 0 \le x_k \lt b$$
Associate to every $a$ in the initial integer interval $[0, b^n]$ the set of ordered pairs
$$\tag 2 \{[k,x_k] \, | \, 0 \le k \lt n \text{ and the base-}b \text{ representation of } a \text{ is given by [1]}\}$$
This association is a
bijective enumeration of $[0, b^n]$ onto the set of all functions
mapping $[0,n-1]$ to $[0,b-1]$.
Since $[0, b^n]$ has $b^n$ elements, we know how to count all the functions from one finite set into another.