Power Structure of Monoid is Monoid

Theorem
Let $\left({G, \circ}\right)$ be a monoid with identity $e$.

Let $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ be the algebraic structure consisting of the power set of $G$ and the operation induced on $\mathcal P \left({S}\right)$ by $\circ$.

Then $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a monoid with identity $\left\{{e}\right\}$.

Proof
By the definition of a monoid, $\left({G, \circ}\right)$ is a semigroup.

By Powerset of Semigroup is Semigroup under Induced Operation, $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a semigroup.

By Subset Product by Identity Singleton, $\left\{{e}\right\}$ is an identity for $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$.

Thus $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$ is a monoid with identity $\left\{{e}\right\}$.