Size of cardinal without choice

Question

How can we show that $ \aleph_0 \leq 2^{2^\kappa}$ for any infinite cardinal $\kappa$ without using the Axiom of Choice?

By Cantor's Theorem we can easily show that if $ \aleph_0 > 2^{2^\kappa}$, then $\aleph_0 > \kappa$. Is there a way to conclude from this that $\kappa$ is finite without appealing to Choice? Thanks

Answer 1

The cardinal $m=2^{2^k} $ is infinite otherwise $k$ must be finite. Therefore $\aleph_0$ must be not greater than $m.$

Answer 2

You can define an injection $f:\mathbb N\to\mathcal P(\mathcal P(X))$ directly: $$ f(n) = \{ A\subseteq X \mid \#A=n \} $$ since, by induction, an infinite set has subsets of every finite cardinality.