Which ordered fields are homeomorphic to their power? It is well known that $\mathbb{R}^2\ncong \mathbb{R}$. It is also known that $\mathbb{Q}^2\cong \mathbb{Q}$. It is a corollary to Sierpiński's theorem which states that every countable metric space without isolated points is homeomorphic to $\mathbb{Q}$. (A proof can be found here and a discussion here.)
As a consequence from the theorem, for every countable subfield $\mathbb{F}\subset \mathbb{R}$ we know that $\mathbb{F}^2\cong \mathbb{F}\cong \mathbb{Q}$.
My question is: what can be said about the general case? In particular, is $\mathbb{R}$ unique in this sense, i.e. 

is it the only ordered field which is not homeomorphic to it's power? 

Putting this in another way, I want know if this property is a completeness axiom (in the sense discussed for example here or here).
 A: $\let\ob\overline\let\sset\subseteq\let\nsset\nsubseteq\DeclareMathOperator\dom{dom}$ There are many such fields different from $\mathbb R$.

Proposition: Let $F$ be a topological field of cardinality $2^\kappa$ with a dense subfield $D$ of cardinality $\kappa$. Then there exists an intermediate field $D\sset K\sset F$ of cardinality $2^\kappa$ such that the powers $K^n$ for $n\in\mathbb N$ are pairwise non-homeomorphic.

Interesting choices include:


*

*$D=\mathbb Q$, $F\subsetneq\mathbb R$. This will produce an incomplete archimedean ordered field $K$.

*$D=\mathbb Q(x)$ with $x>\mathbb Q$, $F=\hat D$ (the Scott completion of $D$, which has cardinality $2^\omega$, being a perfect Polish space). This will produce a nonarchimedean OF of cardinality $2^\omega$.

*Whenever $\kappa^{<\kappa}=\kappa$, there exists an OF $|F|=2^\kappa$ with a dense subfield $|D|=\kappa$. [Under the assumption, there exists a $\kappa$-saturated OF $D$ of cardinality $\kappa$. Then it is easy to construct a nested tree of intervals $\{(a_t,b_t):t\in2^{<\kappa}\}$ such that $b_t-a_t<d_{\mathrm{len}(t)}$, where $\lim_{\alpha\to\kappa}d_\alpha=0$. Any path through the tree defines a Cauchy net, aka good cut, thus the completion $F=\hat D$ has cardinality $2^\kappa$.]
Proof: The main point is that a continuous function $K^n\to K^m$ is uniquely determined by its restriction $D^n\to K^m\sset F^m$. There are only $2^\kappa$ such functions, hence we can diagonalize against them.
So, let $\{f_\alpha,g_\alpha\}_{\alpha<2^\kappa}$ be an enumeration of all pairs of continuous functions $f_\alpha\colon D^{n_\alpha}\to F^{m_\alpha}$, $g\colon D^{m_\alpha}\to F^{n_\alpha}$ where $n_\alpha>m_\alpha$. For any $\alpha$ and $a\in F^{n_\alpha}$, we put
$$\ob f_\alpha(a)=\lim_{\substack{x\in D^{n_\alpha}\\x\to a}}f_\alpha(x)$$
if it exists, and similarly for $\ob g_\alpha$. For $a\in F^{n_\alpha},b\in F^{m_\alpha}$, we define
$$h_\alpha(a)=b\iff \ob f_\alpha(a)=b\text{ and }\ob g_\alpha(b)=a.$$
Note that $h_\alpha$ is a partial injective function. Then, if $D\sset K\sset F$ and $h\colon K^n\to K^m$ is a homeomorphism, there exists an $\alpha<2^\kappa$ such that $h\sset h_\alpha$.
We construct a strictly increasing sequence $\{K_\alpha:\alpha\le2^\kappa\}$ of subfields of $F$, and an increasing sequence $\{A_\alpha:\alpha\le2^\kappa\}$ of subsets of $F$, such that:


*

*$K_\alpha$ and $A_\alpha$ are disjoint, and of cardinality at most $\kappa+|\alpha|$.

*There is no field $K_{\alpha+1}\sset K\sset F$ disjoint from $A_{\alpha+1}$ such that $h_\alpha$ restricts to a homeomorphism $K^{n_\alpha}\to K^{m_\alpha}$.
Then $K=K_{2^\kappa}$ satisfies the required conditions, hence what remains is to carry out the construction. We put $K_0=D$, $A_0=\varnothing$, and for limit $\gamma\le2^\kappa$, we define
$$K_\gamma=\bigcup_{\alpha<\gamma}K_\alpha,\qquad A_\gamma=\bigcup_{\alpha<\gamma}A_\alpha$$
as usual. For the successor step, assume $K_\alpha$ and $A_\alpha$ have been already constructed. We drop the subscripts from $f_\alpha,g_\alpha,h_\alpha,n_\alpha,m_\alpha$ to simplify the notation, and we say that $K$ is a good field if it is a subfield $K_\alpha\sset K\sset F$ disjoint from $A_\alpha$.
Case 1f: There is a good field $K$ such that $K^n\nsset\dom(\ob f)$. We pick $(a_1,\dots,a_n)\in K\smallsetminus\dom(\ob f)$, and define $K_{\alpha+1}=K_\alpha(a_1,\dots,a_n)$, $A_{\alpha+1}=A_\alpha$.
Case 2f: There is a good field $K$ such that $\ob f[K^n]\nsset K^m$. We pick $(a_1,\dots,a_n)\in K^n$ and $i$ such that $b_i\notin K$ for $(b_1,\dots,b_m)=\ob f(a_1,\dots,a_n)$. We put $K_{\alpha+1}=K_\alpha(a_1,\dots,a_n)$, $A_{\alpha+1}=A_\alpha\cup\{b_i\}$.
Cases 1g, 2g: There is a good field such that $K^m\nsset\dom(\ob g)$, or $\ob g[K^m]\nsset K^n$: similar.
Case 3: There is a good field such that $\ob f\restriction K^n$ and $\ob g\restriction K^m$ are not mutually inverse. Say, $a\ne a'\in K^n$, $b\in K^m$, $\ob f(a)=b$, and $\ob g(b)=a'$. We put the coordinates of $a,a',b$ in $K_{\alpha+1}$.
Case 4: None of the previous cases applies, i.e., $h$ restricts to a bijection $K^n\to K^m$ for every good field $K$. I claim that this is in fact impossible. Recall that $n>m$. Since $|K_\alpha(A_\alpha)|<2^\kappa$, we can find $a_1,\dots,a_n\in F$ algebraically independent over $K_\alpha(A_\alpha)$. Then
$$K_\alpha(a_1,\dots,a_n)\cap A_\alpha=K_\alpha\cap A_\alpha=\varnothing,$$
hence $K=K_\alpha(a_1,\dots,a_n)$ is a good field, thus by assumption, $h(a_1,\dots,a_n)=(b_1,\dots,b_m)\in K^m$. Then $K'=K_\alpha(b_1,\dots,b_m)$  has transcendence degree at most $m$ over $K_\alpha$, hence it is a proper subfield of $K$. However, it is also a good field, hence using the assumption again, $h^{-1}(b_1,\dots,b_m)\in(K')^n$, which implies $K\sset K'$, a contradiction.
A minor final complication is that the previous steps do not ensure that $K_\alpha\subsetneq K_{\alpha+1}$. However, this is easy to fix: by the argument in Case 4, there always exists $a\in F$ transcendental over $K_\alpha$ such that $K_\alpha(a)$ is a good field, hence we can throw it in.
