Let $$A = \{1, 2, 3, \ldots, 7\}$$ and let $$P(A)$$ denote the power set of $$A$$. If the number of functions $$f : A \rightarrow P(A)$$ such that $$a \in f(a), \forall a \in A$$ is $$m^n$$, $$m$$ and $$n \in \mathbb{N}$$ and $$m$$ is least, then $$m + n$$ is equal to ______.

For each $$a \in A = \{1, 2, ..., 7\}$$, $$f(a)$$ must be a subset of $$A$$ that contains $$a$$.

For each element $$a$$, the number of subsets of $$A$$ containing $$a$$ is $$2^6 = 64$$ (the other 6 elements can each be in or out).

Since the choices for each element are independent, the total number of functions is $$64^7 = (2^6)^7 = 2^{42}$$.

We need $$m^n = 2^{42}$$ with $$m$$ being the least natural number base. The smallest such $$m$$ is $$m = 2$$, giving $$n = 42$$.

$$m + n = 2 + 42 = \boxed{44}$$.