Is $|\mathbb{R}$| = |$\mathbb{R^2}$| = ... = |$\mathbb{R^\infty}$|? I know that $|\mathbb{R}| = |\mathbb{R^2}|$ because they both contain uncountably many elements, but I find it hard to conceptually understand how $\mathbb{R}$ defined by a line is the same size as $\mathbb{R^2}$ defined by a plane (uncountably many lines), which can be the same size as $\mathbb{R^3}$ defined by uncountably many planes, which is equal to $\mathbb{R^n}$ defined by whatever.
Can someone shed some light on this topic for me?
 A: One can easily create a bijection $f$ from $\mathbb{R}$ to $\mathbb{R}^N$ by using a Z-order curve*, which can be roughly described as the procedure where, if $f(a)=(a_0,a_1,\ldots,a_{n-1})$, then we just read of every $n^{th}$ digit from $a$, starting at the one's place, to yield $a_1$ and do the same, but start at the ten's place to get $a_2$ and, in general, start at the $10^i$'s place to get $a_i$. So, for instance, if $N=2$, we would have $f(\frac{1}{99})=f(.0101\ldots)=(.11\ldots,.00\ldots)=(\frac{1}9,0)$ - to be sure, there is some ambiguity here, in that numbers of the form $\frac{a}{10^b}$ have two representations in decimal, but there are only countably many of those, so we don't need to worry about it - we could also avoid that by interleaving, say, the continued fraction, which is unique.
To be sure, once you know that $|\mathbb{R}|=|\mathbb{R}\times\mathbb{R}|$ anyways, it should be clear that this holds for any $n$ since
$$|\mathbb{R}^n|=|\mathbb{R}^{n-1}\times\mathbb{R}|$$
but, as an inductive hypothesis, we could assume that $|\mathbb{R}^{n-1}|=|\mathbb{R}|$, so
$$|\mathbb{R}^n|=|\mathbb{R}^{n-1}\times\mathbb{R}|=|\mathbb{R}\times\mathbb{R}|=|\mathbb{R}|.$$
I don't think there's any further intuition you can bring to it. Sorry. Cardinality's just like that.
If, by $\mathbb{R}^{\infty}$ you mean the union of every $\mathbb{R}^n$ or equivalently the set of maps $\mathbb{N\rightarrow R}$ (i.e. sequences of reals), then the inductive proof doesn't help there. Unfortunately, the cardinality is no bigger; to prove that, we just do a similar thing to before, but now, we want $f(a)=(a_0,a_1,a_2,\ldots)$ where the sequence continues forever; to do this, one can define:


*

*The $i^{th}$ digit of $a_0$ is the $(2i+1)^{th}$ digit of $a$.

*The $i^{th}$ digit of $a_1$ is the $(4i+2)^{th}$ digit of $a$.

*The $i^{th}$ digit of $a_2$ is the $(8i+4)^{th}$ digit of $a$.


and so on, where the $i^{th}$ digit of $a_n$ is the $(2^{n+1}i+2^n)$ digit of $a$. Since the positions used for $a_0,a_1,a_2,\ldots$ is just the odd numbered digits of $a$, then twice an odd number, then four times and odd number, and so on, no digit of $a$ will be read more than once and every digit will be read once, so this forms a bijection between reals (taken to be series of digits) and sequences of reals.
*Typically, this uses binary, but  I used decimal because it it more familiar to most people.
A: Note first that if saying that "$A$ and $B$ have uncountably many elements" is not enough to say that $|A|=|B|$, as we will see below. You can only say that $|A|>|\mathbb N|$ and $|B|>|\mathbb N|$ (*).
Let's answer to your question now. This is a confusion that most (all?) mathematicians face when studying cardinals for the first time. I think the misconception is about the structure of $\mathbb R$ involved in the problem. The intuitive picture of a set is a messy collection of objects, like a bag containing these objects. Cf. the representation of sets using Venn diagramms (or, as it is also called in France a "potatoes diagram").
But when you think about $\mathbb R$, you never draw it as a potato, but as a tidy, well-organized line. Similar thing happens to $\mathbb R^2$ and $\mathbb R^3$, which are respectively represented as a plane and a space. This is because $\mathbb R$, $\mathbb R^2$, ..., $\mathbb R^n$ are not just sets but sets with structure. The above representations account for the geometric structure of these sets. 
Now, the notion of cardinality forgets about the structures of sets, and just think of them as sets and compare their number of elements. If you want to take the geometry of $\mathbb R^n$ in consideration, then you should see them as vector spaces or manifolds, and compare their dimension.
If it is still difficult to imagine that $|\mathbb R|=|\mathbb R^2|$ (same number of elements), you can imagine the amount of "information" necessary to describe the elements of these sets: forgetting improper representations, a real number $x$ is an infinite sequence of integers given by, for example, its decimal representation $x=a_0,a_1a_2\dots$. A point $(x,y)$ in $\mathbb R^2$ is then given by two sequences $x=a_0,a_1a_2\dots$ and $y=b_0,b_1b_2\dots$ of integers. If you mix these sequences, you obtain a new sequence of integers $a_0b_0,a_1b_1a_2b_2\dots$ (**). Hence, you need only one sequence to describe points in $\mathbb R^2$! It follows that $\mathbb R^2$ does not have more elements than $\mathbb R$.
Finally, by induction, we clearly have that $|\mathbb R^n|=|\mathbb R|$ for any $n\in\mathbb N$.

Notes:
(*) I take "having the same cardinal than $\mathbb N$" for definition of "countable". Some authors would take "being in bijection with a subset of $\mathbb N$" for definition. The latter means that finite sets are also considered as countable.
(**) The $a_0b_0$ at the beginning of the decimal representation is strange. It is actually easier to prove that $|(0,1)^2|=|(0,1)|$ (hence assume that $a_0=b_0=0$) and, since $|R|=|(0,1)|$, this implies that $|\mathbb R^2|=|\mathbb R|$. 
A: Note that:
$$|\mathbb{R}^n|=2^{n\cdot\aleph_0}$$
And
$$|\mathbb{R}^\infty|=2^{\aleph_0\cdot\aleph_0}$$
Also
$$n\cdot\aleph_0=\aleph_0$$
$$\aleph_0\cdot\aleph_0=\aleph_0$$
