Conjecture about ordinal exponentiation: $n^ω=ω$ $\forall n\in\mathbb{N}$-{0,1}

Question

Let $ω$ be the ordinal of the natural numbers. I think this is true:

$n^ω=ω$ $\forall n\in\mathbb{N}-${0,1}

Am i right? If I am wrong, is it true for any $n\in\mathbb{N}$?

Answer 1

Yes. You are correct, if you mean ordinal exponentiation. Recall the definition of ordinal exponentation:

• $\alpha^0=1$,
• $\alpha^{\beta+1}=\alpha^\beta\cdot\alpha$,
• $\alpha^\delta=\sup\{\alpha^\beta\mid\beta<\delta\}$ for limit ordinal $\delta$.

So we have $n^\omega=\sup\{n^k\mid k<\omega\}$. Can you calculate this supremum?