I am trying to calculate $\kappa^\lambda = \aleph\_{\omega\_1}^{\aleph\_0}$. 

I know that if $\kappa$ is a limit cardinal and $0 < \lambda < \mathrm{cf}(\kappa)$ then $\kappa^{\lambda} = \displaystyle \sum\_{\alpha < \kappa} |\alpha|^{\lambda}$.

Hence $ \aleph\_{\omega\_1}^{\aleph\_0} = \displaystyle \sum\_{\alpha < \aleph\_{\omega\_1}} |\alpha|^{\aleph\_0}$. 

I also know that if $\kappa, \lambda$ are infinite cardinals then $(\kappa^+)^\lambda = \kappa^\lambda \cdot \kappa^+$ so that $\aleph\_\alpha^{\aleph\_0} = \aleph\_\alpha \cdot \aleph\_{\alpha-1}^{\aleph\_0} = \aleph\_\alpha \cdot \aleph\_{\alpha-1} \cdot \aleph\_{\alpha-2}^{\aleph\_0} = \dots = \aleph\_\alpha \cdot \aleph\_{\alpha-1} \cdot \dots \cdot \aleph\_{0}^{\aleph\_0} = \aleph\_\alpha$.

Hence $ \aleph\_{\omega\_1}^{\aleph\_0} = \displaystyle \sum\_{\alpha < \aleph\_{\omega\_1}} |\alpha|^{\aleph\_0} = \sum\_{\alpha < \aleph\_{\omega\_1}} |\alpha| = \sum\_{\alpha < \omega\_1} \aleph\_{\alpha}$.

Since for infinite cardinals $\lambda \le \kappa$ we have that $\lambda + \kappa = \kappa$, $\displaystyle \sum\_{\alpha < \omega\_1} \aleph\_{\alpha} = \sup\_{\alpha < \omega\_1} \aleph\_{\alpha} = \aleph\_{\omega\_1}$.

Hence $\aleph\_{\omega\_1}^{\aleph\_0} = \aleph\_{\omega\_1}$. Is this correct? This is an exercise in Just/Weese and the hint is "Assume GCH". I don't think I have used GCH so I suspect I am missing something. Thanks for your help.

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**Edit**

I have used a different method to compute it and have reached the same result (although I still don't know whether this is correct):

By Tarski's theorem, $\aleph\_{\omega\_1}^{\aleph\_0} = \displaystyle \sum\_{\alpha < \aleph\_{\omega\_1}} |\alpha|^{\aleph\_0}$. 

Since $|\alpha|^{\aleph\_0} \le \aleph\_{\omega\_1}$ for all $\alpha < \aleph\_{\omega\_1}$ we get $\displaystyle \sum\_{\alpha < \aleph\_{\omega\_1}} |\alpha|^{\aleph\_0} \leq \aleph\_{\omega\_1}$. Of course, $\aleph\_{\omega\_1} \leq \aleph\_{\omega\_1}^{\aleph\_0}$.

Hence $\aleph\_{\omega\_1} = \aleph\_{\omega\_1}^{\aleph\_0}$.