I'd like someone to check my proof that
$\aleph_\alpha^{\aleph_1} = \aleph_\alpha^{\aleph_0} \cdot 2^{\aleph_1}$ for all $\omega \le \alpha < \omega_1$.
First, let's prove that $\aleph_n^{\aleph_1} = \aleph_n^{\aleph_0} \cdot 2^{\aleph_1}$ for all $n < \omega$ via induction on $n$.
For $n=0$, we have that $\aleph_0^{\aleph_1} = 2^{\aleph_1}$, so the claim holds.
Now, assume the claim for a given $n < \omega$. Then, using the (*) Hausdorff-Formula and the (**) induction hypothesis, we get
$\aleph_{n+1}^{\aleph_1} \overset{(*)}= \aleph_{n+1} \cdot \aleph_n^{\aleph_1} \overset{(**)}= \aleph_{n+1} \cdot \aleph_n^{\aleph_0} \cdot 2^{\aleph_1} \overset{(*)}= \aleph_{n+1}^{\aleph_0} \cdot 2^{\aleph_1}$
as desired.
We now prove the original claim via induction on $\alpha$:
$\begin{eqnarray*}\aleph_\omega^{\aleph_1} = \left(\sup_{n < \omega} \aleph_n \right)^{\aleph_1} \le \left( \prod_{n < \omega} \aleph_n \right)^{\aleph_1} = \prod_{n < \omega} \aleph_n^{\aleph_1} = \prod_{n < \omega} \left( \aleph_n^{\aleph_0} \cdot 2^{\aleph_1} \right) \le \aleph_\omega^{\aleph_0} \cdot 2^{\aleph_1}\end{eqnarray*}$
and if $\alpha = \beta + 1$, then
$\aleph_\alpha^{\aleph_1} \overset{(*)}= \aleph_\alpha \cdot \aleph_\beta^{\aleph_1} \overset{(**)}= \aleph_\alpha \cdot \aleph_\beta^{\aleph_0}\cdot 2^{\aleph_1} \overset{(*)}= \aleph_\alpha^{\aleph_0} \cdot 2^{\aleph_1}$.
Finally, let $\omega < \alpha < \omega_1$ be a limit ordinal and assume the claim below $\alpha$. Then
$\begin{eqnarray*}\aleph_\alpha^{\aleph_1} &=& \left( \sup_{\omega \le \beta < \alpha} \aleph_{\beta}\right)^{\aleph_1} \le \left( \prod_{\omega \le \beta < \alpha} \aleph_{\beta}\right)^{\aleph_1} = \prod_{\omega \le \beta < \alpha} \aleph_{\beta}^{\aleph_1} \overset{(**)}= \prod_{\omega \le \beta < \alpha} \left( \aleph_{\beta}^{\aleph_0}\cdot 2^{\aleph_1} \right) \\ &\overset{\alpha < \omega_1}\le& \aleph_{\alpha}^{\aleph_0} \cdot 2^{\aleph_1} \end{eqnarray*}$