if $\alpha$ is an ordinal is it true that ${\aleph _{\alpha +1}}^{\aleph _{\alpha}}=\aleph _{\alpha +1}$? If we denote the following cardinals: $\beta _0=\aleph _0$,  $\beta _k=2^{\beta _{k-1}}$
then I know that ${\beta _{k+1}}^{\beta _k}=\beta _{k+1}$
but, is it true that for some ordinal $\alpha$,
 ${\aleph _{\alpha+1}}^{\aleph _{\alpha}}={\aleph _{\alpha+1}}$?
I really don't know how to approach this. 

No, that is not necessarily true. In particular, whenever the continuum hypothesis fails so we have $\aleph_1 < 2^{\aleph_0}$, it will still be the case that
$$ \aleph_1^{\aleph_0} \ge 2^{\aleph_0} $$
by simple inclusion.

This doesn't answer whether it is consistent with ZFC that $\aleph_{\alpha+1}{}^{\aleph_\alpha} \ne \aleph_{\alpha+1}$ for all $\alpha$. However, that would follow (by the above argument) if it is consistent that $2^\kappa>\kappa^+$ for all infinite cardinals $\kappa$, which Wikipedia asserts has been proved by Foreman and Woodin, under certain large-cardinal assumptions.
