How to show the following equality of ordinal exponentiation: $n^{\omega^\omega}=\omega^{\omega^\omega}$? Let $n > 1$ be a positive integer. Prove that $n^{\omega^\omega}=\omega^{\omega^\omega}$  
I was able to prove that, since $ω$ is a limit ordinal, and $n > 1$, $$n^\omega= \bigcup_{\beta<\omega}n^\beta = \omega$$
Therefore I end up with $n^{\omega^\omega}=\omega^\omega$, and am left to prove that $\omega^\omega=\omega^{\omega^\omega}$ but that is something that I do not believe holds.
 A: Note that $(n^\omega)^\omega$ is not $n^{\omega^\omega}$. Your calculation was of the former, which is indeed different from $\omega^{\omega^\omega}$.
Let us calculate what $n^{\omega^\omega}$ is. As you know, $\omega^\omega$ is a limit ordinal, so we have that $$n^{\omega^\omega}=\sup\{n^\gamma\mid \gamma<\omega^\omega\}$$
It is enough to calculate this supremum over a cofinal sequence, e.g. $\{\omega^k\mid k\in\omega\}$. So we are suppose to calculate the supremum of $n^{\omega^k}$ for $k\in\omega$. But $\omega^k$ is also a limit ordinal, we once again we unwind the definitions: $$n^{\omega^k} = \sup\{n^\gamma\mid\gamma<\omega^k\}$$
Continuing this unwinding we see that $\omega^k$ is the supremum of $\omega^{k-1}\cdot m$ where $m\in\omega$. But that too is a limit ordinal, so we once again have to find a cofinal sequence $\omega^{k-1}\cdot(m-1)+t$.
So we have that $n^{\omega^1}=\omega$ as you said, and by exponentiation laws we have $n^{\omega^k}=\omega^{\omega^{k-1}}$, but this is really what because now: $$\sup\{n^{\omega^k}\mid k\in\omega\}=\sup\{\omega^{\omega^{k-1}}\mid k\in\omega, k>0\}=\sup\{\omega^{\omega^k}\mid k\in\omega\}=\omega^{\omega^\omega}$$
