Consider cardinality $\alpha = \aleph_0 + 2^{\aleph_0} + 2^{(2^{\aleph_0})}+\dots$, the number of terms is countable. Show that $\alpha$ is the minimal cardinality larger than the cardinalities of sets $\mathbb{N}, \mathcal{P}(\mathbb{N}),\mathcal{P}(\mathcal{P}(\mathbb{N})), \dots.$ Show that $\alpha^{\aleph_0} = 2^\alpha>\alpha$.
I don't know how to solve that and would appreciate help. Here are thoughts I had:
I know that if we add two infinite cardinalities together, the result is equal to the greater cardinality of two; for example, $\aleph_0 + 2^{\aleph_0} = 2^{\aleph_0}$. If we apply this to $\alpha$, we get
.
This $\alpha$ with vertical dots seems to me to be the cardinality of $\mathcal{P}(\mathcal{P}(\mathcal{P}\dots\mathcal{P}(\mathbb{N})\dots))$. However, since the statement must be correct, $\alpha$ should actually be larger than the cardinality of $\mathcal{P}(\mathcal{P}(\mathcal{P}\dots\mathcal{P}(\mathbb{N})\dots))$. Maybe my mistake is that adding two cardinalities together is not the same as adding a countably infinite number of cardinalities together. Or do I misunderstand something fundamentally?
Thank you.