Lower Bound for Ordinal Exponentiation

Theorem
Let $x$ and $y$ be ordinals.

Let $x$ be greater than $1$, where $1$ denotes the successor of the zero ordinal.

Then:


 * $y \le x^y$

Proof
The proof shall proceed by Transfinite Induction on $y$.

Basis for the Induction
If $y = 0$, then $0 \le x^y$ by Empty Set is Subset of All Sets.

This proves the basis for the induction.

Induction Step
The induction hypothesis states that:


 * $y \le x^y$

This proves the induction step.

Limit Case
The induction hypothesis for the limit case states that:


 * $\forall z \in y: z \le x^z$ and $y$ is a limit ordinal

This proves the limit case.