Question about ordinal exponentiation. I am having trouble on how to prove the following identity using ordinal arithmetic. Given a natural number $n > 1$:
$$n^{\omega^{\omega}} = \omega^{\omega^{\omega}}$$
I've tried to use the definition about exponentiation of ordinal numbers when the exponent is an ordinal limit, but I can't reach the result.
Any help will be appreciate it.
 A: For example, we have $n^\omega=\omega$. Now proceeding further we get: $n^{\omega+1}=(n^\omega)\cdot n=\omega \cdot n$. Similarly: $n^{\omega+2}=n^{(\omega+1)+1}=(n^{\omega+1})\cdot n=(\omega \cdot n) \cdot n=\omega \cdot n^2$. Proceeding further: $n^{\omega+3}=n^{(\omega+2)+1}=(n^{\omega+2})\cdot n=(\omega \cdot n^2) \cdot n=\omega \cdot n^3$.
With an induction argument, we could show a general pattern. Now we get:
$n^{\omega \cdot 2}=\sup\{\omega,\omega \cdot n, \omega \cdot n^2, \omega \cdot n^3,.....\}=\omega^2$.
Similarly we would get $n^{\omega \cdot 3}=\omega^3$ and $n^{\omega \cdot 4}=\omega^4$. Following this further, we will get $n^{\omega^2}=\omega^\omega$.

Let's write $f(x)=\omega^x$ and $g(x)=n^x$ generally where $f$ and $g$ are "functions" from ordinals to ordinals. We can show a general equality here I think. That would be $f(N)=g(\omega \cdot N)$ for all ordinals $N \geq 1$. Loosely speaking, the reason is that $1$ step "forward" of the function $f$ is equivalent to $\omega$ steps forward of the function $g$ [ to show this more formally, we would probably use transfinite induction probably, I think ]. That's why this equality works.
(i) Plugging $N=1$ gives $f(1)=g(\omega \cdot 1)=g(\omega)$ which is same as $\omega^1=\omega=n^\omega$.
(ii) Plugging $N=2$ gives $f(2)=g(\omega \cdot 2)$ which is same as $\omega^2=n^{\omega \cdot 2}$.
(iii) Plugging $N=\omega$ gives $f(\omega)=g(\omega \cdot \omega)=g(\omega^2)$ which is same as $\omega^\omega=n^{\omega ^ 2}$.
(iv) Plugging $N=\omega^\omega$ gives $f(\omega^\omega)=g(\omega \cdot \omega^\omega)=g(\omega^\omega)$ which is same as $\omega^{(\omega^\omega)}=\omega^{\omega^\omega}=n^{(\omega ^ \omega)}=n^{\omega ^ \omega}$. This is the required identity.
