Limit Ordinals Closed under Ordinal Exponentiation

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

Let $y$ be a limit ordinal.

Let $x^y$ denote ordinal exponentiation.

Then:


 * If $x > 1$, then $x^y$ is a limit ordinal.


 * If $x \ne \varnothing$, then $y^x$ is a limit ordinal.

Proof
Suppose $x > 1$.

Suppose also that $x^y$ is the successor of some ordinal $w$.

By definition of ordinal multiplication:
 * $\displaystyle x^y = \bigcup_{z \mathop \in y} x^z$

Then:

But $z^+ \in y$ by Successor in Limit Ordinal.

So $w^+ \in x^y$ and $w^+ \in w^+$.

This creates a membership loop and thus is a contradiction by No Membership Loops.

For the second part, since $x$ is not the empty set it follows that $x = z^+$ for some $z$ or that $x$ is a limit ordinal.

If $x$ is a limit ordinal, then $y^x$ is a limit ordinal by the first part.

If $x$ is the successor of $z$, then:


 * $y^x = y^w × y$ by the definition of ordinal exponentiation.

Then, $y^x$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Multiplication.