The cardinal equation $\kappa^{\aleph\_0}=2^\kappa$ is satisfied by $\kappa=\aleph\_0$. 


It is also satisfied by any $\kappa$ for which $MA(\kappa)$ holds. 


Under $GCH$, the equation is satisfied by $\kappa$ if and only if $cof(\kappa)=\aleph\_0$.


So my question is:



> 
> Is it consistent with $ZFC$ that the
>  only solution to
>  $\kappa^{\aleph\_0}=2^\kappa$ is
>  $\kappa=\aleph\_0$?
> 
> 
> 


I´m sorry if this is too basic, but I just don´t see it.



Is there always an uncountable $\kappa$ such that $\kappa^{\aleph_0}=2^\kappa$?
No, it is provable in ZFC that there are many $\kappa$ for which $2^\kappa=\kappa^\omega$. For example, let $\kappa=\beth\_\omega$, and the same argument will work with any strong limit cardinal of cofinality $\omega$. Observe that every subset of $\kappa=\beth\_\omega$ is determined by the sequence of its intersections with all the $\beth\_n$'s. Thus, $2^\kappa\leq \Pi\_n P(\beth\_n)\leq \Pi\_n\beth\_{n+1}\leq\kappa^\omega$. But on the other hand, $\kappa^\omega\leq\kappa^\kappa\leq (2^\kappa)^\kappa=2^\kappa$. So this is a case where $2^\kappa=\kappa^\omega$. 


