proving there is an injection from natural numbers to a set without the axiom of choice Let $\kappa$ be an infinite cardinal number. Prove, without appealing to the Axiom of Choice, that 
  $$\aleph_0 \leq 2^{2^\kappa}$$


I am struggling to answer this question. If I had the axiom of choice available, I would have that there is an injection from natural numbers to k, by cardinal comparability. I'm not sure how to approach this question without axiom of choice being available.
If $S$ is an infinite set, we can define an injection $f:\mathbb N\to\mathcal P(\mathcal P(S))$ by setting
$$f(n)=\{X\subset S:X\text{ is an }n\text{-element set}\}.$$
This shows that $\aleph_0\le2^{2^\kappa},$ a result which I believe is due to Bertrand Russell. Of course it's actually a strict inequality; there is no bijection between $\mathbb N$ and $\mathcal P(\mathcal P(S)).$
By assumption, there exists a non-surjective injective map $f\colon\kappa\to\kappa$.
Define $g_0=f$ and  recursively $g_{n+1}=f\circ g_n$. Finally, define $\Bbb \omega\to 2^{2^\kappa}$, $n\mapsto g_n(\kappa)$.