Cardinality of Set less than Cardinality of Power Set

Theorem
Let $X$ be a set.

Then:
 * $\left\vert{X}\right\vert < \left\vert{\mathcal P \left({X}\right)}\right\vert$

where
 * $\left\vert{X}\right\vert$ denotes the cardinality of $X$,
 * $\mathcal P \left({X}\right)$ denotes the power set of $X$.

Proof
By No Bijection from Set to its Power Set:
 * there exist no bijections $X \to \mathcal P \left({X}\right)$

Then by definition of set equivalence:
 * $X \not\sim \mathcal P \left({X}\right)$

Hence by definition of cardinality:
 * $(1): \quad \left\vert{X}\right\vert \ne \left\vert{\mathcal P \left({X}\right)}\right\vert$

By Cardinality of Set of Singletons:
 * $(2): \quad \left\vert{\left\{{\left\{{x}\right\}: x \in X}\right\}}\right\vert = \left\vert{X}\right\vert$

By definition of subset:
 * $\forall x \in X: \left\{{x}\right\} \subseteq X$

Then by definition of power set:
 * $\forall x \in X: \left\{{x}\right\} \in \mathcal P \left({X}\right)$

Hence by definition of subset:
 * $\left\{{\left\{{x}\right\}: x \in X}\right\} \subseteq \mathcal P \left({X}\right)$

Then by Subset implies Cardinal Inequality and $(2)$:
 * $\left\vert{X}\right\vert \leq \left\vert{\mathcal P \left({X}\right)}\right\vert$

Thus by $(1)$:
 * $\left\vert{X}\right\vert < \left\vert{\mathcal P \left({X}\right)}\right\vert$