 Solution

The recursive ordinal addition definitions are $a+0=a$, $a+(b+1)=(a+b)+1$, and $a+\lambda ={\sup }_{b<\lambda }(a+b)$ for a nonzero limit ordinal. Ordinal multiplication is $a\cdot 0=0$, $a\cdot (b+1)=a\cdot b+a$, and $a\cdot \lambda ={\sup }_{b<\lambda }a\cdot b$. For $a>0$, ordinal exponentiation is $a^0=1$, $a^{b+1}=a^b\cdot a$, and $a^{\lambda }={\sup }_{b<\lambda }{a}^b$; the exceptional zero base has $0^0=1$ and $0^b=0$ for $b>0$.

Transfinite induction gives ${\omega }^a\ge a$. At zero this is immediate; at a successor, ${\omega }^{b+1}={\omega }^b\omega \ge {\omega }^b+1\ge b+1$. At a limit, continuity and the induction hypothesis give ${\sup }_{b<a}{\omega }^b\ge {\sup }_{b<a}b=a$. Also $a\mapsto {\omega }^a$ is strictly increasing: its successor values strictly increase, and continuity preserves the strict inequality between any two distinct indices.

For $\alpha >0$, ${\omega }^{\alpha +1}>{\omega }^{\alpha }\ge \alpha$, so there is a least $\delta$ with ${\omega }^{\delta }>\alpha$. It cannot be zero and cannot be a limit: at a limit, a supremum greater than $\alpha$ has a term already greater than $\alpha$, contradicting minimality. Thus $\delta ={\alpha }_0+1$, andStrict increase proves uniqueness, and ${\alpha }_0\le {\omega }^{{\alpha }_0}\le \alpha$ proves the requested index bound.

Choose the largest positive integer $a_0$ with ${\omega }^{{\alpha }_0}{a}_0\le \alpha$. It exists because $\alpha <{\omega }^{{\alpha }_0}\omega ={\sup }_{m<\omega }{\omega }^{{\alpha }_0}m$. The remaining tail of the ordinal interval has an order type $\rho$ with $\alpha ={\omega }^{{\alpha }_0}{a}_0+\rho$; maximality implies $\rho <{\omega }^{{\alpha }_0}$. If $\rho >0$, repeat the construction on it. The leading exponents strictly decrease, and there is no infinite strictly decreasing sequence of ordinals: its set of values would have a least member followed by a smaller one. Hence the process terminates, yielding the Cantor normal formThe first exponent is determined by the preceding band, the first coefficient by maximality, and the tail by the ordinal interval left after that initial segment. Recursion therefore proves uniqueness of every exponent and coefficient. The PDF's second coefficient is printed ${\alpha }_1$ rather than $a_1$; the stated natural-number coefficients and the construction identify the intended expression.

For the final assertion, powers of $\omega$ are additively indecomposable: if $\tau <{\omega }^{\eta }$, then $\tau +{\omega }^{\eta }={\omega }^{\eta }$ for $\eta >0$. One proves this by induction on $\eta$. At a successor, bound $\tau$ by a finite multiple of ${\omega }^{\eta -1}$ and use the supremum over all such multiples; at a limit, bound $\tau$ by a smaller power and use continuity. For $\eta =0$ the only possible $\tau$ is zero. Thus a smaller leading exponent is absorbed when a larger-leading ordinal is added on its right. If $\beta ,\gamma$ have unequal leading exponents, either $\beta +\gamma =\gamma$ or $\gamma +\beta =\beta$. If their leading exponents coincide, the leading coefficient of either sum is the sum of their two positive leading coefficients; it exceeds each coefficient, so neither absorption equality holds. Zero is never commensurable with an ordinal. Consequently commensurability is exactly equality of the leading exponent, equivalently membership of both positive ordinals in one of the stated bands.