Archimedean completeness of some fields

I need a reference (different from Hahn's 1907 paper) for the following result.

Theorem: If $G$ is a totally ordered abelian group, then the field $\mathbb{R}((G))$ is archimedean complete.

• $\mathbb{R}((G))$ consists of all the functions $f:G\to\mathbb{R}$ such that $\{g\in G:f(g)\neq0\}$ is well-ordered.
• Let $E$ be an ordered field. Two nonzero elements $x,y\in E$ are comparable if there exist $m,n\in\mathbb{N}$ such that $|x|<m|y|$ and $|y|<n|x|$ (where $|a|$ is defined as $\max\{a,-a\}$).
• Let $E/ K$ be an extension of ordered fields, where the order on $E$ restricted to $K$ coincides with that of $K$. We say that $E$ is an Archimedean extension of $K$ if for every $x\in E$, there exists $y\in K$ such that $x$ and $y$ are comparable in $E$.
• A field $K$ is Archimedean complete if it has no proper archimedean extension fields.

Does anyone know a good reference for the proof of this result? The more the better.

There are many such proofs in the literature. The completeness for Hahn fields follows from the completeness for Hahn groups, since every Hahn field is a Hahn group. For a simple proof of the latter, see pp. 862-863 of Note on Hahn's Theorem on Ordered Abelian Groups, A.H. Clifford, Proc. Am Math. Soc. 5 (1954), pp. 860-863. Many other references can be found in the following question of mine on Mathoverflo Hahn's Embedding Theorem and the oldest open question in set theory.

• I understand the embedding theorem, but in the references there is not mention about the notion of Archimedean completeness. If $K$ is an archimedean extension of $\mathbb{R}((G))$, then $K$ is of type $G$ (the group of Archimedean classes of $K$ is isomorphic to $G$). Thus, by the embedding theorem there exists an embedding $\theta$ from $K$ to $\mathbb{R}((G))$. But still can happen that $K$ is a proper extension of $\mathbb{R}((G))$. How can we discard that possibility? In general, this situation is possible. For example $K((x))$ can be embedded in $K((x^2))$ being a proper extension of it – Chilote Nov 3 '16 at 22:10
• @Chilote. You are mistaken. Look at the last paragraph on p. 862. It begins: "Theorem 3.2 (i.e. Hahn's Embedding Theorem) is beautifully adapted to the proof of Hahn's completeness theorem.... Hahn defines a complete ordered group to be an ordered group $G$ which does not admit a proper extension $H$ which is Archimedean relative to $G$....." – Philip Ehrlich Nov 4 '16 at 1:57
• I have to admit I overlooked big part of that page. Thanks for the keen reference. – Chilote Nov 4 '16 at 5:22