A summary of the relevant cardinal arithmetic, by the way (in the presence of choice): $$~$\aleph_{\alpha} + \aleph_{\alpha} = \aleph_{\alpha} = \aleph_{\alpha} \aleph_{\alpha}$~$$ while $$~$2^{\aleph_{\alpha}} > \aleph_{\alpha}$~$$
by Patrick Stevens Jul 9 2016
A summary of the relevant cardinal arithmetic, by the way (in the presence of choice): $$~$\aleph_{\alpha} + \aleph_{\alpha} = \aleph_{\alpha} = \aleph_{\alpha} \aleph_{\alpha}$~$$ while $$~$2^{\aleph_{\alpha}} > \aleph_{\alpha}$~$$