There are $n!$ such functions if $A=\{ 1,...,n\}$. Let $f(A)=:a_1$. Then also $f(B)=a_1$ for all $B\ni a_1$ because $A=B\cup(A\backslash B)$ and $A\backslash B$ doesn't contain $a_1$. Now let $f(A\backslash\{a_1\})=:a_2$. Then similarly, $f(B)=a_2$ for all $B\ni a_2, B\not\ni a_1$. ... Let $f(A\backslash\{a_1,...,a_{k-1}\})=:a_k$. Then $f(B)=a_k$ for all $B\ni a_k, B\not\ni a_1,...,a_{k-1}$. ... Carry on until $k=n$. Now the $a_1,...,a_n$ are all different by construction, so they form a permutation of $\{ 1,...,n\}$. And it is easy to see that each nonempty $B\subset A$ is captured exactly once by the above 'filtering'. So there is a 1-1 correspondence between the functions and the permutations of $\{ 1,...,n\}$.