I was studying Cantor's contributions to set theory, and I came upon two results:

#1: It can be shown that the power set of the naturals $\displaystyle 2^{\mathbb{N}}$ has a higher cardinality than the naturals $\displaystyle \mathbb{N}$ by attempting to enumerate all the subsets of $\displaystyle \mathbb{N}$ and being able to create a new subset previously not listed, thus contradicting the fact that there was originally a bjiection between $\displaystyle 2^{\mathbb{N}}$ and $\displaystyle \mathbb{N}$.

#2: It can also been shown that the reals $\displaystyle \mathbb{R}$ are "more numerous" than the naturals $\displaystyle \mathbb{N}$ by using Cantor's Diagonalization argument.

Thus, it is known that both $\displaystyle 2^{\mathbb{N}}$ and $\displaystyle \mathbb{R}$ have a higher cardinality than the naturals $\displaystyle \mathbb{N}$.

But how can we show that $\displaystyle 2^{\mathbb{N}}$ and $\displaystyle \mathbb{R}$ have the same cardinality as each other, though? I tried constructing a bijection (because I know that one must exist); I tried to use Cantor's style; nothing really worked.

I am so stuck on this proof that I do not even know where to start. Someone give me a hint or a starting point, please?