Count of Binary Operations on Set

Theorem
Let $$S$$ be a set whose cardinality is $$n$$.

The number $$N$$ of possible different binary operations that can be applied to $$S$$ is given by:


 * $$N = n^{\left({n^2}\right)}$$

Proof
A binary operation on $$S$$ is by definition a mapping from the cartesian product $$S \times S$$ to the set $$S$$.

Thus we are looking to evaluate:
 * $$N = \left|{\left\{{f: S \times S \to S}\right\}}\right|$$

The domain of $$f$$ has $$n^2$$ elements, from Cardinality of Cartesian Product.

The result follows from Cardinality of Set of All Mappings.

Comment
The number grows rapidly with $$n$$:

$$\begin{array} {c|rr} n & n^2 & n^{\left({n^2}\right)}\\ \hline 1 & 1 & 1 \\ 2 & 4 & 16 \\ 3 & 9 & 19 \ 683 \\ 4 & 16 & 4 \ 294 \ 967 \ 296 \\ \end{array}$$

We're still only at 4 elements in a set, and we're already up to 4 thousand million different possible algebraic structures.