For an infinite cardinal $\kappa$, $\aleph_0 \leq 2^{2^\kappa}$ I'm trying to do a past paper question which states:
$$
\text{For all infinite cardinals $\kappa$, we have } \aleph_0 \leq  2^{2^\kappa}.
$$
I'm supposed to be able to do this without the axiom of choice, but I can't see how. I know that there can be no bijection (or surjection) from any natural number with $X$, where $|X| = \kappa$, but I can't get any further.
A solution or hint would be great, thanks! 
EDIT: Something that bugs me about this question is that they've written $\aleph_0 \leq  2^{2^\kappa}$, whereas the solution below using Cantor's theorem shows that $\aleph_0 <  2^{2^\kappa}$, that is strict inequality holds. Is this just a mistake, or to throw people off?

HINT: It suffices to find a surjection from $2^\kappa$ onto the natural numbers, now look at cardinalities of finite sets. 
You can also show that equality is never possible. But that is besides the point. You are asked to find an injection, not to prove there are no bijections. 
