I attempted to prove that there's a surjective function $\mathbb{N} ^{|2^\mathbb{N}|} \rightarrow 2^\mathbb{N}$ and that there isn't one in the opposite way, however I wasn't able to prove the former.
Is there another way to prove this cardinal inequality?
That is quite easy to see. Clearly $\mathbb N^{2^\mathbb N}\geq 2^{2^\mathbb N}$ (as any function that maps into $\{0,1\}$ also maps into $\mathbb N$). But as $|2^A| = |\mathcal P(A)|$ for any set, and by a simple proof by cantor $\mathcal P(A)>A$ for any $A$, we get $$ 2^{2^\mathbb N} > {2^\mathbb N}$$ and thus your statement.