Ramsey property and linear orders on $\kappa$ I have been trying to solve to prove the following statement:

Let $\kappa$ be an uncountable cardinal. The following are equivalent:

*

*Every linear order of cardinality $\kappa$ has a suborder of
order-type $\kappa$ or $\kappa^*$ ($\kappa$ inverted).

*$\kappa\longrightarrow (\kappa)_2^2$

The fact that (2) implies (1) is easy to show. For (1) implies (2), I have been trying to define a linear order based on an arbitrary coloring on $[\kappa]^2$, but so far my idea does not work.
Can anybody give me some suggestion?
 A: I am writing this answer so that the question is removed from the unanswered queue, but it is likely that there is a clever, more direct proof.$\newcommand{\funcs}[2]{{}^{#1}{#2}}$$\newcommand{\lelex}{<_{\mathrm{lex}}}$

I am going to show that (1) implies that $\kappa$ is inaccessible and has the tree property. It is a standard fact that these two properties are equivalent to (2) (see e.g. Theorem 7.8 in Kanamori's book).
First, $\kappa$ is regular. Otherwise, suppose that $\langle\kappa_\alpha;\alpha<\lambda\rangle$ for some $\lambda<\kappa$ is a cofinal sequence in $\kappa$. Consider the linear order $L=\sum_{\alpha<\lambda}[\kappa_\alpha,\kappa_{\alpha+1})^*$, that is, start with $\lambda$ and replace each $\alpha\in\lambda$ with the interval $[\kappa_\alpha,\kappa_{\alpha+1})$ ordered in reverse. Any increasing sequence in $L$ can have only finitely many elements in each block, and there are not enough blocks for an increasing sequence of length $\kappa$ to exist. On the other hand, any decreasing sequence in $L$ must eventually be contained in a single block, and none of the blocks are long enough for a decreasing sequence of length $\kappa$ to exist. But this contradicts (1).
Secondly, $\kappa$ is a strong limit. Otherwise, suppose that there is $\lambda<\kappa$ such that $2^\lambda\geq\kappa$. Consider the linear order $L=(\funcs{\lambda}{2},\lelex)$. Suppose that there is an increasing or decreasing sequence $\langle f_\alpha;\alpha<\kappa\rangle$ in $L$. In either case note that if $\alpha<\beta<\gamma$ then $f_\gamma$ splits away from $f_\alpha$ no later than $f_\beta$ splits away from $f_\alpha$. Now fix $f_0$. By the previous observation there are a node $t_0\in \funcs{<\lambda}{2}$ on the branch $f_0$ and an ordinal $\alpha_0<\kappa$ such that for every $\alpha\geq\alpha_0$ the branch $f_\alpha$ splits from $f_0$ at $t_0$. Now repeat this argument with $f_{\alpha_0}$ to get a new node $t_1\sqsupset t_0$ and a tail of the branches that split from $f_{\alpha_0}$ at $t_1$. Repeat this $\lambda$ many times and observe that the nodes $t_\alpha$ now form a branch $g\in \funcs{\lambda}{2}$. But $\kappa$ is regular, so the sequence of ordinals $\langle\alpha_\xi;\xi<\lambda\rangle$ is bounded by some $\alpha_\lambda<\kappa$. But from the way we constructed the branch $g$, it follows that all the branches after $f_{\alpha_\lambda}$ are equal to $g$, leading to a contradiction.
Thirdly, $\kappa$ has the tree property. Let $T$ be a $\kappa$-tree; we may assume that it is a subtree of $\funcs{<\kappa}{\kappa}$. Consider the linear order $L=(T,\lelex)$, where $s\lelex t$ if $s$ is an initial segment of $t$ or, when they split, $s$ splits to the left.
Suppose there is a lex-decreasing sequence of nodes $\langle t_\alpha;\alpha<\kappa\rangle$. As before, there is a node $s_0\sqsubset t_0$ such that a tail of the sequence splits from $t_0$ at $s_0$. Furthermore, there is an ordinal $\eta_0$ such that a (possibly smaller) tail of the sequence passes through $s_0^\frown \eta_0$. Now repeat this argument $\kappa$ many times as before to build a branch through $T$.
Now suppose that there is a lex-increasing sequence of nodes $\langle t_\alpha;\alpha<\kappa\rangle$. It may be the case that this sequence contains a branch, in which case we would be done. So assume there is no cofinal subsequence that would form a branch. Let $P_0$ be a maximal linearly ordered (in the tree order) subset of the $t_\alpha$ containing $t_0$. By our assumption a tail of the $t_\alpha$ is not contained in $P_0$, so each $t_\alpha$ in this tail must split away from $\bigcup P_0$. Since, again, later nodes in the sequence must split away no later than earlier nodes, a tail of them must split away from $\bigcup P_0$ at the same node $s_0$. But since $T$ is a $\kappa$-tree, there must be an ordinal $\eta_0$ such that a (possibly smaller) tail of the sequence passes through $s_0^\frown \eta_0$. Now repeat this argument $\kappa$ many times to build a branch through $T$ along the nodes $s_\alpha$.
