Let's assume $2^{\aleph_3}=\aleph_4$ and $\left(\aleph_{\omega_1}\right)^{\aleph_1}\neq\left(\aleph_{\omega_1}\right)^{\aleph_2}$.
Prove that $$\exists_{\alpha\in Lim}\left( \left(\aleph_{\alpha}\right)^{\aleph_2} > \aleph_{\omega_1}\ \wedge\ \forall_{\gamma <\alpha}\ \left(\aleph_{\gamma}\right)^{\aleph_2} \leq \aleph_{\omega_1}\right).$$
It's a question I found in one of the Set Theory exams at my Uni. To be honest, I have absolutely no idea where to start.
Could you give me a helping hand?
Since successor cardinals are regular, if $\kappa^+>\lambda$ then any function from $\lambda$ to $\kappa^+$ is bounded, thus ${}^\lambda\kappa^+=\bigcup_{\alpha<\kappa^+}{}^\lambda\alpha$, and therefore $$ (\kappa^+)^\lambda\le\sum_{\alpha<\kappa^+}|\alpha|^\lambda=\kappa^+ \cdot\kappa^\lambda. $$ By the way, this arguments goes back to Hausdorff. Note that we in fact have $$(\kappa^+)^\lambda=\kappa^+\cdot\kappa^\lambda,$$ since the left hand side is obviously larger than or equal to the right hand side. Also, if $\kappa^+\le\lambda$, we also have equality since both sides equal $2^\lambda$.
In particular, if $\tau$ is a limit cardinal larger than $\kappa$, and $\kappa^\lambda\le\tau$, then also $(\kappa^+)^\lambda\le\tau$. It follows that the first $\mu$ such that $\mu^\lambda>\tau$ is either a limit cardinal or less than or equal to $\lambda$.
To apply the above to the case at hand, where $\tau=\aleph_{\omega_1}$ and $\lambda=\aleph_2$, note that the first of your assumptions ensures that if $\kappa^\lambda>\tau$, then $\kappa>\aleph_2$ since ${\aleph_2}^{\aleph_2}\le{\aleph_3}^{\aleph_3}=2^{\aleph_3}$.