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.