Is $ \infty^{\infty} = \aleph_0$?

Question

Note: I read something about this on the internet somewhere once, my logic could be 100% flawed.

Is $(\aleph_0)^{\infty} = \infty^{\infty}$?

Suppose the number $\infty^{\infty}$. Assuming $\infty$ in this sense has a cardinality of $\aleph_0$, could it be represented as $(\aleph_0)^{\infty}$?

As an extension, because $\aleph_0$ has a cardinality in it's set (some sort of countable infinities), is $(\aleph_0)^n = \aleph_0$?

It seems to be that $\infty^{\infty} > \infty$, and by extension, $\infty^{\infty} > \aleph_0$, but at the same time, if $\aleph_0$ is the $1$ of it's own set, $(\aleph_0)^\infty = \aleph_0$ the same way $1^1 = 1$.

Answer 1

$\aleph_0$ is the cardinal number of the set of naturals. If you are working in set theory it is well defined. $\infty$ is an informal concept usually used when you are using the reals, the integers, or something like that. It is not a number in any of those systems but used as something to take limits to, for example.

In set theory $\aleph_0^{\aleph_0}$ makes sense. It is the cardinality of the set of functions $\Bbb N \to \Bbb N$, for example. It is equal to $\mathfrak c$, the cardinality of $\Bbb R$. As $\infty$ is informal, I don't know what to make of $\infty^{\infty}$

Answer 2

I have read the following chain of cardinal numbers in a Topology book

$1<2<3<......<\aleph_0<2^{\aleph_0}=c< 2^c < 2^{2^c}<.....$

Clearly, this chain has only one kind of countable infinity i.e. $\aleph_0$ and beyond this there is an infinite hierarchy of uncountable infinities which are all distinct from onf another.