Consider the set $A \times A$ which has $n^2$ elements. The requirement that your relation $R$ be reflexive says that $(a,a) \in R$ for all $a$. So out of the $n^2$ elements of $A \times A$, $n$ of them are required to be in $R$. What remains is $n^2-n$ elements of $A \times A$, each of which may or may not be in the relation. The subsets of these $n^2-n$ elements correspond one-to-one with all reflexive binary relations, and there are $2^{n^2-n}$ such subsets.
You should prove two statements:
1) If $A$ has $m$ elements and $B$ has $n$ elements, then $A \times B$ has $mn$ elements.
2) If $C$ is a set of $k$ elements, then the power set $P(C)$ has $2^k$ elements.
Regarding 2): Let $C = \{c_1,\ldots,c_k\}$ and consider the mapping
$$f : P(C) \to \{ (a_1,\ldots, a_k) ~ | ~ a_i \in \{0,1\}\}$$
defined in the following way: For $D\subseteq C$ let $f(D) := (d_1,\ldots, d_k)$ where $d_i = 1$ if $c_i \in D$ and $d_i =0$ if $c_i \notin D$. For example $f(\emptyset) = (0,\ldots, 0)$, $f(\{c_2\}) = (0,1,0,\ldots, 0)$ and so on.
This way we associate every subset $D$ of $C$ with a $k$-tuple that has a $1$ at the $i$-th position, if $c_i$ is in $D$ and a $0$ otherwise.
Now one can see that $f$ is a bijection: If $f(D) = f(D')$, then $D = D'$ and for every sequence $(a_1,\ldots, a_k)$ consisting of $0$'s and $1$'s we can find a matching subset $D \subseteq C$ such that $f(D) = (a_1,\ldots, a_n)$. That $f$ is a bijection means that $P(C)$ and the set of all possible $k$-tuples containing only $0$'s and $1$'s have the same cardinality (number of elements).
Another hint to finish the proof of 2):
$$\{ (a_1,\ldots, a_k) ~ | ~ a_i \in \{0,1\}\} = \underbrace{\{0,1\} \times \ldots \times \{0,1\}}_{k \text{ times}}$$
Can you use 1) now?
I hope this isn't too confusing. If you have further questions, just leave a comment below.
Best Answer
Techniquely a relation should be a relation over the sets $X_k$. Even though $\in$ and $\subset$ are called "relations", they aren't really defined in this way. They simply come from definitions and axioms of set theory.
Of course you could try to see if you could define $\in$ and $\subset$ over some sets $X_k$, but then you immediately see that you somehow need "the universe of objects" or "the set of all sets", which aren't sets.