Ordinal Exponentiation is Closed

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

Then:


 * $x^y \in \operatorname{On}$

That is, ordinal exponentiation is closed.

Proof
Let $x = 0$ and $y = 0$.

Then by the definition of ordinal exponentiation:
 * $x^0 = 1$

Let $x = 0$ and $y \ne 0$.

Then by the definition of ordinal exponentiation:
 * $x^0 = 0$

In either case, $x^y$ is an ordinal.

Now suppose that $x \ne 0$.

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

Basis for the Induction
By definition of ordinal exponentiation:
 * $x^0 = 1$

which is an ordinal.

This proves the basis for the induction.

Induction Step
The inductive hypothesis states that $x^y \in \operatorname{On}$.

Suppose the inductive hypothesis holds.

Then:

This proves the induction step.

Limit Case
The inductive hypothesis states that $x^z \in \operatorname{On}$ for all $z \in y$.

This proves the limit case.