Set of all Self-Maps under Composition forms Monoid

Theorem
Let $S$ be a set.

Let $S^S$ be the set of all mappings from $S$ to itself.

Let the operation $\circ$ represent composition of mappings.

Then the algebraic structure $\left({S^S, \circ}\right)$ is a monoid whose identity element is the identity mapping on $S$.

Proof
By Set of All Self-Maps is Semigroup, $\left({S^S, \circ}\right)$ is a semigroup.

By Identity Mapping is Left Identity and Identity Mapping is Right Identity the identity mapping on $S$ is the identity element of $\left({S^S, \circ}\right)$.

Since $\left({S^S, \circ}\right)$ is a semigroup with an identity element, it is a monoid by definition.