Why are the cardinality of $\mathbb{R^n}$ and $\mathbb{R}$ the same? As I study the first part of abstract algebra, I have a question: why $\left\vert{\mathbb R}\right\vert = \left\vert{\mathbb R^2}\right\vert$?
And moreover, I knew the fact that $\left\vert{\mathbb R}\right\vert = \left\vert{\mathbb R^n}\right\vert$.
Does the equality apply to any infinite set?
How can I prove them?
 A: You can think of a real number in decimal notation: an infinite sequence of digits with a decimal point somewhere in there. Now you can take $n$ (e.g. $n=3$) such numbers and interleave their digits like this:
      1  2.  3  5  6…
  1  3  7 . 2  7  3 …
    9  3  .1  5  8  …
⇒ 10931372.123575836…

So this way you have turned a triple of real numbers into a single real number in a reversible fashion. To make this rigorous you'd have to consider special cases like $0.999… = 1$, but the above should convey the general idea in an intuitive way.
For $n=\infty$ this argument no longer holds in this fashion. And it's not that trivial there, either, since it depends on the kind of infinity. Most likely you have contable infinity in mind there, I guess, so you are asking about $\mathfrak c^{\aleph_0}$.
A: The existence of space-filling curves shows that the cardinality of $\mathbb{R}^n$ can be at most that of $\mathbb{R}$. A space-filling curve in a curve parametrized by a surjective mapping $s: \mathbb{R} \rightarrow \mathbb{R}^n$. Since $s$ is surjective, we must conclude that $|\mathbb{R}^n| \leq |\mathbb{R}|$.
More generally, to show that we have $|A^n| = |A|$ for any infinite set $A$ and finite $n \geq 1$, it is enough to show that $|A \times A| = |A|$, since the general case then follows by a trivial inductive argument.
To prove the result, consider the family $\mathcal{F}$ of those functions $f$ such that $f: X \times X \rightarrow X$, where $X$ is a subset of $A$. $\mathcal{F}$ is non-empty, for if $X$ is a countable subset of $A$, then the usual dovetailing argument tells us that such an $f$ exist. It is also easy to see that $\mathcal{F}$ can be equipped with a partial order $\sqsubseteq$, namely that of extension: $f_1 \sqsubseteq f_2$ if $\mathrm{dom}(f_1) \subseteq \mathrm{dom}(f_2)$ and whenever $f_1(x) = y$, then also $f_1(x) = y$. Because $\mathcal{F}$ is partially ordered, we know from Zorn's lemma that it has a maximal element $f_{max}$. So let $\mathrm{dom}(f_{max}) = X$. All that is left is then to show that $|X| = |A|$. 
Suppose $X$ had strictly smaller cardinality than $A$. But then, since $|A| = \max(|X|,|A \setminus X|)$ we must have that $|A| = |A \setminus X|$ and therefore that $|X| < |A \setminus X|$. This then means that we can find a $Y \subset A \setminus X$ of the same cardinality as $X$. But now consider the sets $X \times Y$, $Y \times X$ and $Y \times Y$. Each of these sets is infinite and has the same cardinality as $X \times X$ and therefore also the same cardinality as $X$ and as $Y$. Therefore we have 
$$ |X \times Y \cup Y \times X \cup Y \times Y| = |Y|$$
But $X \times Y \cup Y \times X \cup Y \times Y = (X \cup Y) \times (X \cup Y)$, so this means that we can extend $f_{max}$ such that its domain is $(X \cup Y) \times (X \cup Y)$, and that is a contradiction, since $f_{max}$ was assumed to be maximal.
