I'm trying to determine if $\aleph_{\alpha+1}^{\aleph_a} > \aleph_\alpha^{\aleph_a}$ is true for some ordinal $\alpha$. I don't think $\aleph_{\alpha+1} > \aleph_\alpha$ necessarily implies it just because of the way $\aleph$'s work. Or can I just assume that?
Any advice would be appreciated! Thank you.
Recall that $$2^\kappa \le \kappa^{\kappa} \le (2^{\kappa})^\kappa = 2^{\kappa \times \kappa} = 2^{\kappa}$$ so equality holds throughout and so $\kappa^\kappa = 2^\kappa$.
And notice that $$2^{\aleph_\alpha} \le \aleph_{\alpha+1}^{\aleph_\alpha} \le (2^{\aleph_\alpha})^{\aleph_\alpha}$$
where the latter comes from noting that $2^{\aleph_\alpha}$ is bigger than $\aleph_\alpha$, but $\aleph_{\alpha+1}$ is the smallest cardinal bigger than $\aleph_\alpha$. So again equality holds throughout, because $(2^{\aleph_\alpha})^{\aleph_\alpha} = 2^{\aleph_\alpha}$.