$$\omega\rightarrow(\omega)^n_k$$ is proven by induction on $n$ .

If $n=1$ then this just states that any partition of an infinite set into a finite number of subsets must include an infinite set; that is, the union of a finite number of finite sets is finite. This is simple enough to prove: since there are a finite number of sets, there is a largest set of size $x$ . Let the number of sets be $y$ . Then the size of the union is no more than $xy$ .

If $$\omega\rightarrow(\omega)^n_k$$

then we can show that

$$\omega\rightarrow(\omega)^{n+1}_k$$

Let $f$ be some coloring of $[S]^{n+1}$ by $k$ where $S$ is an infinite subset of $\omega$ . Observe that, given an $x<\omega$ , we can define $f^x\colon[S\setminus\{x\}]^{n}\rightarrow k$ by $f^x(X)=f(\{x\}\cup X)$ . Since $S$ is infinite, by the induction hypothesis this will have an infinite homogeneous set.

Then we define a sequence of integers $\langle n_i\rangle_{i\in\omega}$ and a sequence of infinite subsets of $\omega$ , $\langle S_i\rangle_{i\in\omega}$ by induction. Let $n_0=0$ and let $S_0=\omega$ . Given $n_i$ and $S_i$ for $i\leq j$ we can define $S_{j}$ as an infinite homogeneous set for $f^{n_i}\colon[S_{j-1}]^n\rightarrow k$ and $n_j$ as the least element of $S_j$ .

Obviously $N=\bigcup \{n_i\}$ is infinite, and it is also homogeneous, since each $n_i$ is contained in $S_j$ for each $j\leq i$ .