Cardinal arithmetic and diagonalization.
The cardinality of the power set is $2^{|X|}$, which is why we write the power set as $2^X$, which also represents the set of characteristic functions of subsets of $X$ (i.e. the set of functions $X\to\{0,1\})$.
This is very cool when you first realize it: A function of the form above is determined by its elements. Given such a function $f: X\to\{0,1\}$, let $S_f:=f^{-1}(1)$ be the subset defined by $f$, and given a subset $S\subseteq X$, let $f_S$ denote the characteristic function of $S$ with respect to $X$. Then we see immediately that these functions $S_{(-)}:\{0,1\}^X\to \mathcal{P}(X)$ and $f_{(-)}: \mathcal{P}(X)\to \{0,1\}^X$ are inverse to one another. Therefore, it follows that the power set is precisely a function set as noted above. Edit: I should note that in abstract nonsense language, this means that that $2=\{0,1\}$ is the "subobject classifier" for the category of sets.
To see that $|\mathbf{R}^\mathbf{N}|=|\mathbf{R}|$, notice that this becomes $(2^\mathbf{N})^\mathbf{N}\cong 2^{\mathbf{N}\times \mathbf{N}}\cong 2^\mathbf{N}\cong \mathbf{R}$, where $\cong$ here represents an isomorphism of sets, or a bijective function (baby Rudin calls this equivalence relation "equipotence").
To find that $|\mathbf{R}|=2^{|\mathbf{N}|}$, this is a matter of using the technique called diagonalization.
To find that the integers and rationals represent the same cardinal as $\mathbf{N}$, you have to do a little bit of tricky encoding, but it is easy to find on the page about cardinality on wikipedia.
To see a worked out proof of why $|2^X|>|X|$ for any set $X$, look at this whimsical link.