Proving that $(\omega_n)^\omega=\omega_n$ providing CH but not GCH

HINT: Use Hausdorff's formula which states that $\aleph_{\alpha+1}^{\aleph_\beta}=\aleph_\alpha^{\aleph_\beta}\cdot\aleph_{\alpha+1}$. And induction.

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 $n$.