Every ordinal number $α$ can be uniquely written as ${\displaystyle \omega ^{\beta _{1}}c_{1}+\omega ^{\beta _{2}}c_{2}+\cdots +\omega ^{\beta _{k}}c_{k}}$, where $k$ is a natural number, ${\displaystyle c_{1},c_{2},\ldots ,c_{k}}$ are positive integers, and ${\displaystyle \beta _{1}>\beta _{2}>\ldots >\beta _{k}\geq 0}$ are ordinal numbers. This decomposition of $α$ is called the Cantor normal form of $α$, and can be considered the base-$ω$ positional numeral system. The highest exponent ${\displaystyle \beta _{1}}$ is called the degree of ${\displaystyle \alpha }$ , and satisfies ${\displaystyle \beta _{1}\leq \alpha }$ . The equality ${\displaystyle \beta _{1}=\alpha }$ applies if and only if ${\displaystyle \alpha =\omega ^{\alpha }}$. In that case Cantor normal form does not express the ordinal in terms of smaller ones; this can happen as explained below.
A minor variation of Cantor normal form, which is usually slightly easier to work with, is to set all the numbers $c_i$ equal to 1 and allow the exponents to be equal. In other words, every ordinal number $α$ can be uniquely written as ${\displaystyle \omega ^{\beta _{1}}+\omega ^{\beta _{2}}+\cdots +\omega ^{\beta _{k}}}$, where k is a natural number, and ${\displaystyle \beta _{1}\geq \beta _{2}\geq \ldots \geq \beta _{k}\geq 0}$ are ordinal numbers.
My question is how to calculate the Cantor normal form of the ordinal $α^α$ where $α$ is the ordinal $(ω^{ω^{2}} + 7)ω^{ω^{3}}ω + ω + ω4$.
Is it possible to use theorem in the following link?
https://proofwiki.org/wiki/Ordinal_Exponentiation_via_Cantor_Normal_Form/Limit_Exponents
Thank you for your time and help.
This is not a complete answer yet.
First let's consider the ordinal: $(\omega^{\omega^{2}} + 7)\omega^{\omega^{3}}\omega + \omega + \omega 4$. The first thing we can note is that this expression is well-defined because addition and multiplication (of ordinals) is associative. First, let us simply this expression in various steps.
(1) $(\omega^{\omega^{2}} + 7)\omega^{\omega^{3}} \cdot \omega + \omega + \omega \cdot 4$
(2) $(\omega^{\omega^{2}} + 7)\omega^{\omega^{3}} \cdot \omega^1 + (\omega + \omega \cdot 4)$
(3) $(\omega^{\omega^{2}} + 7)\omega^{\omega^{3}+1} + \omega \cdot 5$
(4) $(\omega^{\omega^{2}})\omega^{\omega^{3}+1} + \omega \cdot 5$
The reason for (3) to (4) is that if $l$ is a limit ordinal then $(\alpha+7) \cdot l=\alpha \cdot l$ (for any arbitrary $\alpha \geq 1$).
(5) $\omega^{\omega^{2}+(\omega^{3}+1)}+ \omega \cdot 5$
(6) $\omega^{\omega^{3}+1}+ \omega \cdot 5$
The reason for (5) to (6) is that $\omega^{2}+(\omega^{3}+1)=(\omega^{2}+\omega^{3})+1=\omega^3+1$
So essentially we want $m^m$ where $m=\omega^{\omega^{3}+1}+ \omega \cdot 5$.
Edit1: If what is written below is correct (I need to check the details though), then it should take one pretty close to the answer. In the next edit I will try to complete it.
So we define $m_0=\omega^{\omega^{3}}$ and $m_1=\omega^{\omega^{3}+1}$. Let's first try to calculate ${m_0}^{m_0}$. I think we can show that for any arbitrary $\alpha \geq 1$, we should have: ${m_0}^\alpha={\omega}^{\omega^3 \cdot \alpha}$. If we add the additional condition that $\alpha=\omega^\omega \cdot n$ (for some ordinal $n \geq 1$), then we have $\omega^3 \cdot \alpha =\alpha$. However, $\omega^{\omega^{3}}=m_0$ satisfies the condition in last sentence. Hence $\omega^3 \cdot m_0 =m_0$.
Due to last paragraph, we can write ${m_0}^{m_0}={\omega}^{\omega^3 \cdot m_0}=\omega^{m_0}=\omega^{(\omega^{\omega^3})}=\omega^{\omega^{\omega^3}}$. As an aside we see that $m_0^{(\omega^\omega)}={\omega}^{(\omega^3 \cdot {\omega^\omega})}={\omega}^{({\omega^\omega})}=\omega^{\omega^\omega}$. This is not too different from $(\omega^\omega)^{(\omega^\omega)}=\omega^{\omega^\omega}$.
Now let's come to calculating ${m_1}^{m_0}$. I think we can show that for any arbitrary $\alpha \geq 1$, we should have: ${m_0}^\alpha={\omega}^{(\omega^3+1) \cdot \alpha}$. If we add the additional condition that $\alpha=\omega^\omega \cdot n$ (for some ordinal $n \geq 1$), then we have $(\omega^3+1) \cdot \alpha =\alpha$. However, $\omega^{\omega^{3}}=m_0$ satisfies the condition in last sentence. Hence $(\omega^3+1) \cdot m_0 =m_0$.
Due to the last paragraph we get ${m_1}^{m_0}={\omega}^{(\omega^3+1) \cdot m_0}=\omega^{m_0}=\omega^{(\omega^{\omega^3})}=\omega^{\omega^{\omega^3}}$.
Now let's focus our attention $m^{m_0}$ where $m=\omega^{\omega^{3}+1}+ \omega \cdot 5$. It seems to me that there might be an easier way to deduce it based on something like this. Define $m_2=\omega^{\omega^{3}+2}$. Now we should have $m<m_2$.
So we have $m_1<m<m_2$. Hence we must have $m_1^{m_0} \leq m^{m_0} \leq m_2^{m_0}$. But it seems like ${m_2}^{m_0}=\omega^{\omega^{\omega^3}}$ based on reasoning similar to ${m_1}^{m_0}=\omega^{\omega^{\omega^3}}$. The inequality mentioned above seems to force us to conclude that $m^{m_0}=\omega^{\omega^{ \omega^3}}$.
If we have an expression for $m^{m_0}$ then it seems that getting $m^m$ from there shouldn't be too difficult.
(To be completed)