Proving that $(\omega_n)^\omega=\omega_n$ providing CH but not GCH This is an exercise from a book from Kunen - SET THEORY, An Introduction to Independence Proofs
Assume CH but don't assume GCH. Show that $(\omega_n)^\omega=\omega_n$ for $1 \le n < \omega$.
I don't have a clue for a good starting point...
 A: HINT: Use Hausdorff's formula which states that $\aleph_{\alpha+1}^{\aleph_\beta}=\aleph_\alpha^{\aleph_\beta}\cdot\aleph_{\alpha+1}$. And induction.
A: Recall that $\omega_1$ is a regular cardinal. Now, consider $(\omega_1)^\omega$ and let $A \in (\omega_1)^\omega$. Define the $sup(A) = \alpha$ such that $\alpha$ is the smallest ordinal where every $ \beta \in A$, $ \beta < \alpha$.   Since $cof(\omega_1) = \omega_1$, we note that $\alpha$ must always be a countable ordinal. Partition $ (\omega_1)^\omega$ into equivalence classes where $A$~ $B$ if $sup(A) = sup(B)$. Denote the equivalence class for some ordinal $\alpha$ and $O_\alpha$. Notice that $|O_\alpha| = 2^{\aleph_0} = \aleph_1 $ for $\omega<\alpha <\omega_1$ (and $|O_\alpha| = \emptyset$ for any finite $\alpha$).  
Thus, $(|\omega_1)^\omega| = |\bigcup_{\omega<\alpha<\omega_1} O_\alpha|$. But the RHS is just the $\omega_1$ union of sets of size $\omega_1$ (which has cardinality $\omega_1$). 
Now you can finish the problem by induction on $n$. 
A: By CH, we know that $2^\omega = \omega_1$. Thus, ${\omega_1}^\omega = (2^\omega)^\omega = 2^{\omega\cdot\omega} = 2^\omega = \omega_1$.
Now assume that ${\omega_n}^\omega = \omega_n$. Because $\omega_{n+1}$ is regular, any map from $\omega$ to $\omega_{n+1}$ cannot be cofinal.  Thus the set of maps from $\omega$ to $\omega_{n+1}$ is the union of the set of maps from $\omega$ to $\alpha$ where the cardinality of $\alpha$ is $\omega_n$.  By our inductive hypothesis, each element of this union has size $\omega_n$ and there are $\omega_{n+1}$ members of this union, so the set of all maps from $\omega$ to $\omega_{n+1}$ has size no greater than $\omega_{n+1}$.  Q.E.D.
