Define $\beth_n(\kappa)$ inductively by $\beth_0(\kappa)=\kappa$ and $\beth_{n+1}(\kappa)=2^{\beth_n(\kappa)}$. Erdős-Rado theorem says $\beth_n(\kappa)^+\rightarrow(\kappa^+)^{n+1}_\kappa$ holds, i.e., any coloring of the complete $n+1$-hypergraph on $\beth_n(\kappa)^+$ many vertices using $\kappa$ many colors has a homogeneous set of size $\kappa^+$. The case $n=1$ says any edge coloring of the complete graph on $(2^\kappa)^+$ many vertices has a homogeneous set.
I have a fairly clear picture of what's going on in the $n=1$ case. Basically we copy the proof of infinite Ramsey theorem, but in order to be less wasteful we build a tree instead of a sequence. As starting point let $0$ be the root of the tree we are going to construct. $(2^\kappa)^+\setminus\{0\}$ is partitioned into $\kappa$ many pieces $(A_{(i)}:i<\kappa)$, by the color of the edge connecting that point to $0$. Pick $a_{(i)}\in A_{(i)}$, say the least element, then $A_{(i)}\setminus\{a_{(i)}\}$ is further partitioned into $(A_{(ij)}:j<\kappa)$ by the color of the edge connecting that point to $a_{(i)}$. Continue indefinitely, and at limit stages take intersection along all the branches (as long as the result is nonempty); if $A_s$ is a singleton it does not have children. We thus get a (Hausdorff) tree in which every node has at most $\kappa$ many children, with the crucial property that for each node, the edges between it and its descendents all have the same color. Since at the end there are $(2^\kappa)^+$ many nodes, a bit of calculation shows the height must be at least $\kappa^++1$, so there is a branch of length $\kappa^+$, and some subset of that branch would be the desired homogeneous set.
Is my following understanding of the case $n=2$ correct? Let $\lambda=\beth_2(\kappa)^+$. We let $0$ be the root, and $1$ be its only child. $\lambda\setminus\{0,1\}$ is partitioned into $\kappa$ many pieces. From each piece $A$ choose a point $a$. Next we want to partition $A\setminus\{a\}$ into subpieces so that not only $\{1,a,b\}$ all have the same color for $b$ from the same subpiece $B$, but the same holds for $\{0,1,b\}$ and $\{0,a,b\}$. There are thus at most $\kappa^3=\kappa$ many subpieces. In general, a node at height $\alpha$ has at most $\kappa^{\tau}$ many children, where $\tau$ is the number of $2$-subsets of $\alpha$, namely $|\alpha|$ as long as $\alpha$ is infinite. This tree has height at least $(2^\kappa)^++1$, and we can apply the case $n=1$ to get a homogeneous set.
Is this argument correct? I'm not sure if it produces (essentially) the same tree as lemma 7.2 in Kanamori's The Higher Infinite. Kanamori requires that if $\xi$ has descendent $\eta$ then $\xi<\eta$, though it doesn't seem necessary to me. Also he adds $\eta$ to the tree by placing it on top of a chain $b$ that is maximal with respect to the property that $\{\xi_1,\dots,\xi_n\}$ and $\{\xi_1,\dots,\xi_{n-1},\eta\}$ have the same color whenever $\xi_1<\cdots<\xi_n$ are in $b$; is such a $b$ necessarily maximal in the tree constructed thus far? If not it seems rather different from my argument.
Another short question: does the tree argument also work to show that $\kappa^+\rightarrow(\kappa)^2_2$ for singular strong limit $\kappa$? Strangely I've never seen this presumably equally important case in standard textbooks.