In Harry Gonshor's *An Introduction to the Theory of Surreal Numbers*, on page 50, Gonshor points to a method for intuitively guessing what the square root of the countable infinity is in his construction -- I have a generalization of this intuitive guess method that I am curious about. 


For some context, in Gonshor's construction:



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> A **surreal number** is a function $x:\alpha\rightarrow\{+,-\}$ from an ordinal $\alpha\in O\_n$ into a two element set $\{+,-\}$, where we refer to $\alpha$ as the *length* of $x$ and write it as $\ell\_x$ in general. The class $\mathbb{R}\_\infty$ of all such functions is called the **Surreal numbers**, and it is ordered lexicographically by $$-<0<+,$$ where by convention $x(\ell\_x)=0$ for all $x$; that is, a surreal number takes on a value of $0$ at the first ordinal for which it is undefined, its length.
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I believe the above is sufficient for the discussion at hand, although I will add a definition of multiplication if it is deemed appropriate. The positive natural number $n$ is given by the string of $n$ $+$'s, and the negative integer $-n$ is given by a string of $n$ $-$'s, and in general we can obtain a positive or negative ordinal $\alpha$ or $-\alpha$ as a string of $\alpha$ $+$'s or $-$'s respectively. 


At the first successor position where the sign changes we begin averaging, so $$++=2$$ $$+++=3$$ $$+++-=\frac{2+3}{2}=2.5$$ $$+++--=\frac{2+2.5}{2}=2.25$$ $$+++-+=\frac{2.5+3}{2}=2.75$$ $$\dots$$ in very similar fashion to binary expansions of real numbers with $1\equiv+$ and $0\equiv-$ (note that we would have to pick a new symbol for the thing ordered between $+$ and $-$).


If we look at a string of $\omega$ $+$'s followed by $\omega$ $-$'s, we end up with $\frac{\omega}{2}$: $$\frac{\omega}{2}=+++\dots---\dots,$$ and I believe in general we have that for [$\gamma$-numbers](https://en.wikipedia.org/wiki/Additively_indecomposable_ordinal) $\alpha$ a string of $\alpha$ $+$'s followed by $\alpha$ $-$'s is $\frac{\alpha}{2}$, and strings of $+$'s and $-$'s which switch at limit ordinals all generate nice ordinal fractions in this manner as long as the strings are of additively comparable length, much like an iterated version of the averaging process above. 


More interestingly, if we take a string of $\omega$ $+$'s and then attach a string of $\omega^2$ $-$'s we get the square root of $\omega$; that is, $$\sqrt{\omega}=+++\dots---\dots---\dots---\dots.,$$ where $\dots.$ indicates that we repeat this countable string of $-$'s a countable number of times. I suspect we then have that $\omega$ $+$'s followed by a string of $\omega^3$ $-$'s is the cube root of $\omega$: $$\sqrt[3]{\omega}=+++\dots---\dots.---\dots.,$$ and we can obtain higher finite roots in the expected fashion. My question is this:



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> Is a string of $\omega$ $+$'s followed by $\omega^\omega$ $-$'s the $\omega^{th}$ root of $\omega$? That is, does $$\sqrt[\omega]{\omega}=+++\dots---\dots..$$
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If yes, my followup questions are:


1. Does this behavior generalize in the expected fashion to higher transfinite roots, or to higher transfinite $\gamma$-numbers?
2. Does this imply that there is a canonical 'infinite product' structure on $\mathbb{R}\_\infty$, seeing as we have canonical infinite-th roots?


Any assistance is greatly appreciated.



Roots of $\omega$, larger $\gamma$-numbers
Chapter 3 of ONAG (see parts "$\omega$-map" and "Sign-expansions and normal forms") contains a way to obtain $\omega^x$ for any number $x$ (including transfinite cardinals). **UPD (replacing a false statement)**: according to the reference, for $x$ having sign expansion $x\_0 x\_1 \ldots$ (indexed by ordinals up to some $\alpha$), the sign expansion of $\omega^x$ is $x\_0^{e\_0 + 1} x\_1^{e\_1 + 1} \ldots$, where $e\_\beta$ is the (ordinal) number of $+$'s among $x\_{\delta}$ for $\delta < \beta$. According to this, the number $+^{\omega}-^{\omega^3}$ is $\sqrt[\omega]{\omega}$ rather than $\sqrt[3]{\omega} = +^{\omega}-^{\omega^2}+^{\omega^2}-^{\omega^3}+^{\omega^3}\ldots$ ($\omega$ terms).


In particular case when $x$ is a limit ordinal we can see that $x = +^x$ and $1/x = + -^x$. It follows that $\sqrt[x]{\omega} = +^{\omega} -^{\omega^2 x}$, and $+^{\omega} -^{\omega^{\omega}} = \sqrt[\omega^{\omega}]{\omega}$. 


