Count of Binary Operations with Identity

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

The number $$N$$ of possible different binary operations which have an identity element that can be applied to $$S$$ is given by:


 * $$N = n^{\left({n-1}\right)^2 + 1}$$

Proof
From Count of Binary Operations with Fixed Identity, there are $$n^{\left({n-1}\right)^2}$$ such binary operations for each individual element of $$S$$.

As Identity is Unique, if $$x$$ is the identity, no other element can also be an identity.

As there are $$n$$ different ways of choosing such an identity, there are $$n \times n^{\left({n-1}\right)^2}$$ different algebraic structures with an identity.

These are guaranteed not to overlap by the uniqueness of the identity.

Hence the result.

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

$$\begin{array} {c|cr} n & \left({n-1}\right)^2 + 1 & n^{\left({n-1}\right)^2 + 1}\\ \hline 1 & 1 & 1 \\ 2 & 2 & 4 \\ 3 & 5 & 243 \\ 4 & 10 & 1 \ 048 \ 576 \\ \end{array}$$