Simplifying $\left( 2^{\aleph_\alpha}\right)^{\aleph_0}$ Let $X = \omega_\lambda^\omega$ be the product space where each $\omega_\lambda$ has the discrete topology. I'm able to bound the cardinality of open sets in $X$ below by $2^{\aleph_\lambda}$. The bound above should also be $2^{\aleph_\lambda}$. But it comes down to proving that $$\left(2^{\aleph_\lambda}\right)^{\aleph_0} = 2^{\aleph_\lambda}$$
Is this result immediate as the exponent would be $\aleph_\lambda \cdot \aleph_0 = \text{max}(\aleph_\lambda, \aleph_0) = \aleph_\lambda $? I'm not extremely familiar with cardinal exponentiation, so I'm not sure if I'm able to make such a simplification (bringing the exponent inside). If that's true, is there a good reference as to why I'm able to do that?

Yes. Keep in mind that $|A|^{|B|} = |A^B|$ by definition. Now in any Cartesian Closed Category, we have the following isomorphism, natural in $A$:
$\begin{equation}
\begin{split}
Hom(A, (B^C)^D)
&\simeq Hom(D \times A, B^C)\\
&\simeq Hom(C \times (D \times A), B) \\
&\simeq Hom((C \times D) \times A, B) \\
&\simeq Hom(A, B^{C \times D})
\end{split}
\end{equation}$
so by the Yoneda Lemma, we have an isomorpism $(B^C)^D \simeq B^{C \times D}$.
Thus, we have for any sets $B, C, D$, we have $(|B|^{|C|})^{|D|} = |B^C|^{|D|} = |(B^C)^D| = |B^{C \times D}| = |B|^{|C \times D|} = |B|^{|C| \cdot |D|}$.
Thus, for any cardinals $\alpha, \beta, \gamma$, we have $(\alpha^\beta)^\gamma = \alpha ^ {\beta \cdot \gamma}$.
So in particular, we have $(2^{\aleph_\alpha})^{\aleph_0} = 2^{\aleph_\alpha \cdot \aleph_0}$. And since $\alpha \geq 0$, we have $\aleph_\alpha \cdot \aleph_0 = \aleph_\alpha$. So the cardinal simplifies to $2^{\aleph_\alpha}$.
