Proof

The set of all functions $2^{\mathbb N}=\{f\mid \mathbb N\to \{0,1\}\}$ is equipotent to the power set $\mathcal P(\mathbb N)$, since every function mapping all natural numbers to the set $\{0,1\}$ is an indicator function (see also explanation about the connection between the power set and mapping to two elements)
Thus, $|\mathcal P(\mathbb N)|=|2^{\mathbb N}|.$
On the other hand, there is no surjective function of a set to its power set, and therefore neither a surjective function.
Trivially, there is an injective function $f:\mathbb N\to\mathcal P(\mathbb N)$ (just set $f(a)=\{a\}$ for all $a\in\mathbb N.$)
Comparing both cardinal numbers yields $|\mathbb N| < \mathcal P(\mathbb N)=2^{\mathbb N}.$
Thus, $2^{\mathbb N}$ is uncountable.

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