Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC? In complex analysis, there is a function called Euler's Gamma function. Whenever given a positive integer $n+1$, it will return $n!=\prod_{i=1}^{i < n+1}i$.
I'm not sure if there is similar function for infinite cardinals such that
$$\Gamma(\aleph_\alpha)=\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa$$,
but at least we can evaluate the value of that production.
So my question: Is $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$?
 A: The study of infinite products is very subtle in general. The "factorial" is reasonably well understood. In 1925, Tarski proved that $\prod_{\xi<\beta}\aleph_\xi=(\aleph_{\bigcup \beta})^{|\bigcup\beta|}$ (for details, see this blog post of mine). Since $\prod_{n\in\omega}n=2^{\aleph_0}$, we get that 
 $$ \Gamma(\aleph_\alpha)=2^{\aleph_0}\prod_{\xi<\alpha}\aleph_\xi=2^{\aleph_0}(\aleph_{\bigcup\alpha})^{|\bigcup\alpha|}. $$
(Whether or not this is $\beth_\alpha$ is independent of $\mathsf{ZFC}$ in general, and clearly depends on the size of the two powers that appear in this displayed expression. As suggested in the comments, it may be worth pointing out that, for example, if $\alpha=\tau+1$ where $\tau$ is a countable successor ordinal and $2^{\aleph_0}<\aleph_\tau$, then $\Gamma(\aleph_\alpha)=2^{\aleph_0}\aleph_\tau^{|\tau|}=\aleph_\tau^{\aleph_0}=\aleph_\tau<2^{\aleph_\tau}\le2^{\aleph_\alpha}$.)
A: (Note: This answer was given to a previous revision of the question)
Certainly not. Consider a model of $\sf ZFC$ such that for every countable ordinal, $2^{\aleph_\alpha}=\aleph_{\omega_1+1}=\beth_1$.
Let $\alpha$ be a countable ordinal and consider the following, $$\prod_{\beta<\alpha}\aleph_\beta\leq\prod_{\beta<\alpha}\aleph_\alpha = \aleph_\alpha^{\aleph_\alpha}=2^{\aleph_\alpha}=\beth_1\ll\beth_\alpha.$$
