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I shall explain why I'm asking this because it may seems something tricky.

We know what's $\Bbb R^{\Bbb R}$, it's the set of all functions from real numbers to real numbers, whose cardinal is $\aleph_2$.

Another way to intepret this set: It's the set of all functions which turns points into points. It's trivial sentence.

Now, let's see the set $(\Bbb R^{\Bbb R})^\Bbb R$, it's the set of functions from real numbers to functions from real numbers to real numbers (not equal to $\Bbb R^{\Bbb R^{\Bbb R}}$, this is the set of functions from real-to-real functions to real numbers), whose cardinal is $\aleph_3$.

Another way to interpret this set: It's the set of all functions which turns points into functions.

For instance: $$f:\Bbb R \to \Bbb R^\Bbb R,\; f(x)=g:\Bbb R\to\Bbb R ,\;y\mapsto g(y)=sin(y)+x$$ $f \in (\Bbb R^{\Bbb R})^\Bbb R$.

Analogously, we define $((\Bbb R^{\Bbb R})^\Bbb R)^\Bbb R$, whose cardinal is $\aleph_4$.

Are the elements of these sets some kind of functions that "spread" their domain points? I mean, the previous $f$ spreads $x$ into the points in $g(y)=sin(y)+x$ graph.

So, the last question: Does a function in $((\Bbb R^{\Bbb R})^\Bbb R)^\Bbb R$ spread points two times?

For instance, $h \in ((\Bbb R^{\Bbb R})^\Bbb R)^\Bbb R$

$$h:\Bbb R \to (\Bbb R^{\Bbb R})^\Bbb R \\x\mapsto h(x)=f:\Bbb R \to \Bbb R^\Bbb R,\\ y \mapsto f(y)=g:\Bbb R\to\Bbb R ,\\z\mapsto g(z)=z^2·sin(y)+x.$$

If I'm correct interpreting these, I think I have good examples of sets with cardinal $\aleph_n, n \in \Bbb N$.

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    I think that for arbitrary sets $\mathrm{E}^{\mathrm{A} \times \mathrm{B}}$ is canonically indentified with $(\mathrm{E}^\mathrm{A})^\mathrm{B}$ via the function $f:(a, b) \mapsto f(a,b)$ goes to $\tilde{f}$ given by $\tilde{f}(b):a \mapsto f(a, b).$ So, the space you want to ask about would be just $\mathbb{R}^{\mathbb{R}^3}$ the functions from vectors in $\mathbb{R}^3$ to $\mathbb{R}.$ – 2017-01-14
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    "whose cardinal is $\aleph_2$" **This is incorrect**. The cardinality of $\mathbb{R}^\mathbb{R}$ is that of $2^{2^{\aleph_0}}$, which is **not necessarily the same as** $\aleph_2$. The statement $2^{\aleph_\alpha}=\aleph_{\alpha+1}$ is the *Generalized Continuum Hypothesis*, and is independent of ZFC; even specific instances of it, like $2^{\aleph_0}=\aleph_1$ (the *Continuum Hypothesis*) and $2^{\aleph_1}=\aleph_2$, are independent of ZFC. In fact, even really "obvious" facts - like $\kappa>\lambda$ implies $2^\kappa>2^\lambda$ - are independent of ZFC! Care must be taken . . . – 2017-01-14
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    It might be a good idea to revise the definitions of $\aleph$ numbers and infinite cardinals in general. It is not true that taking any power will immediately increase the cardinality of your set. – 2017-01-14
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    I've justed analyzed what you all answered and now I see those sets I've defined are equivalente to multi variable functions with real numbers as images. So, what could be examples of set with cardinal $\aleph_3$ or $\aleph_4$ ? – 2017-01-14

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Currying (or uncurrying, to be more precise) provides a bijection between $(\Bbb R^{\Bbb R})^{\Bbb R})^\cdots)^\Bbb R$ and $\Bbb R^{\Bbb R^n}$ (where $n$ is the numbers of $\Bbb R$'s appearing as an exponent). In particular: if $f \in (\Bbb R^{\Bbb R})^{\Bbb R})^\cdots)^\Bbb R$, then we can define $\Phi(f) \in \Bbb R^{\Bbb R^n}$ by $$ \Phi(f)(x_1,\dots,x_n) = [[[f(x_1)](x_2)]\cdots](x_n) $$ With the appropriate axioms (I think the axiom of choice is enough), it can be shown that $\Bbb R^n \sim \Bbb R$, whereby we deduce that $$ (\Bbb R^{\Bbb R})^{\Bbb R})^\cdots)^\Bbb R \sim \Bbb R^{\Bbb R^n} \sim \Bbb R^\Bbb R $$ That is, all of the sets that you're looking at have the same cardinality as $\Bbb R^\Bbb R$ (which you assume in your question to be $\aleph_2$).

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    Please correct your statement concerning $\aleph_2$ – 2017-01-14
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    Specifically, Pedro's point is that you're assuming the GCH: in ZFC alone, $2^{2^{\aleph_0}}$ is **not necessarily** the same as $\aleph_2$. – 2017-01-14
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    @NoahSchweber Are you answering something? So I stop doing it – 2017-01-14
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    @Noah thanks for clarifying – 2017-01-14
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    @NoahSchweber Sorry for writing so quickly. I was asking if your writing an answer to the question. I'm doing it now, but if you were also, I'd stop writing mine. – 2017-01-14
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    @PedroSánchezTerraf No, I'm not - I was just clarifying your comment to the OP. – 2017-01-14
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This is some sense supplemental to the other answer (and repeats various comments since I started writing it before they were there ;-).

The cardinal $\aleph_2$ is the third infinite cardinal, and as such it's the second uncountable cardinal. It can't be proved on the basis of current axioms which of those $\aleph_\alpha$ is the cardinality of $\mathbb{R}$. Since there is a natural bijection between $\mathbb{R}$ and $\{0,1\}^{\mathbb{N}}$, the notation for the cardinal of $\mathbb{R}$ is $2^{\aleph_0}$. And actually this notation satisfies many pleasant properties of usual exponentiation. On the other hand, we have that for any two sets $A,B$, $|A^B|=|A|^{|B|}$. So we can calculate as follows: $$ |\mathbb{R}^{\mathbb{R}}| = |\mathbb{R}|^{|\mathbb{R}|} = (2^{\aleph_0})^{2^{\aleph_0}} = 2^{\aleph_0 \cdot 2^{\aleph_0}} = 2^{2^{\aleph_0}}, $$ where in the last equality I used the fact that $\kappa\cdot\lambda =\max \{\kappa,\lambda\}$ for every pair of infinite cardinals $\kappa$ and $\lambda$.