The task is:
Show that $$ \langle (x_n)_{ n \in \mathbb{N}} \mid x_n^n = x_{n-1} \text{ for } 1 < n \in \mathbb{N} \rangle $$
is presentation of additive group $(\mathbb{Q},+)$.
Can you explain me how to show this? Or explain simple what is presentation?
See Wikipedia for an introduction to group presentations. The traditional definition using an explicit construction and elements is very concrete, but it has a major drawback: Often you cannot describe the (normal form) of the elements of a group defined by a presentation (since the involved normal closures are horrible), and often you don't want to deal with elements at all, but with homomorphisms. Objects are not really interesting for themselves; mathematics starts when you compare these objects via morphisms. For this it is useful to know universal properties of your objects. These basically just give a description of the morphisms from or into your object.
I don't think that you can solve this exercise with the traditional definition of a group presentation without making cumbersome calculations (see the last paragraph). These calculations are redundant, in fact not necessary at all when you work with the universal properties instead. So let me explain the universal property of group presentations (which is not mentioned at Wikipedia, sic!), which I actually see as the conceptual definition of group presentations. As a matter of fact we don't really need groups here, the same definition applies to any type of algebraic structure. In order to simplify the notation, I assume that the generators and the relations are countable.
If $x_1,x_2,\dotsc$ are symbols and $R_1,R_2,\dotsc$ are group-relations in these symbols (for example $R_1 = (x_3^2 x_2^{-1} x_1^{-5} = x_4^3)$), then $$\langle x_1,x_2,\dotsc \mid R_1,R_2,\dotsc \rangle$$ is defined to be a group satisfying the following universal property:
In category-theoretic language, this means that $\langle x_1,x_2,\dotsc \mid R_1,R_2,\dotsc \rangle$ represents the functor $\mathsf{Grp} \to \mathsf{Set}$ which maps $G$ to the set of all families of elements $(g_1,g_2,\dotsc)$ in $G$ satisfying the relations $R_1(g),R_2(g),\dotsc$ in $G$.
For example, $\langle x \mid x^n=1 \rangle$ represents the functor which maps a group $G$ to its $n$-torsion $\{g \in G : g^n=1\}$. Or even more simple, $\langle x \mid - \rangle$ represents the functor which maps $G$ to its underlying set $|G|$. You can show that it is given by $\mathbb{Z}$, since $\hom(\mathbb{Z},G) \cong |G|$ via $f \mapsto f(1)$. Then, the homomorphism theorem (which describes homomorphisms on quotients) yields $\hom(\mathbb{Z}/n\mathbb{Z},G) \cong \{g \in G : g^n=1\}$, i.e. $\langle x \mid x^n = 1 \rangle = \mathbb{Z}/n\mathbb{Z}$. See Wikipedia for more examples.
Now let us describe $\hom(\mathbb{Q},-)$. Notice that $\mathbb{Q}$ is generated by the sequence of elements $\frac{1}{n}$ ($n>0$) as an abelian group, but their relations are a little bit complicated. So let us choose instead $x_n := \frac{1}{n!}$ ($n \geq 0$), here we simply have $n \cdot x_n = x_{n-1}$. Because of $(n-1)! \cdot x_n = \frac{1}{n}$ these $x_n$ still generate $\mathbb{Q}$. If $G$ is a group, written additively (although not assumed to be commutative), then a homomorphism $f : \mathbb{Q} \to G$ thus is determined by the values $f(x_n)$ for $n \geq 0$ and these values satisfy the relations $n \cdot f(x_n) = f(x_{n-1})$ for $n \geq 1$. Also notice that $f(\frac{p}{n}) = p (n-1)! f(x_n)$, this describes $f$ completely.
We claim that the converse is also true: If $G$ is a group containing elements $g_0,g_1,\dotsc$ such that $n \cdot g_n = g_{n-1}$, then there is a unique homomorphism $f : \mathbb{Q} \to G$ such $f(x_n)=g_n$. Well, we have to define $f(\frac{p}{n}) = p (n-1)! g_n$ for $p \in \mathbb{Z}$ and $n \in \mathbb{N}$, $n>0$. One checks that $f$ is well-defined and a homomorphism. This requires some calculation, but it is not hard. This will finish the proof that $\mathbb{Q}$ satisfies the defining universal property of the group presentation (now written multiplicatively again) $$\langle x_0,x_1,\dotsc \mid \forall n \geq 1 : x_n^n = x_{n-1} \rangle.$$
But we don't need to do these calculations when we use an even more abstract but still very easy approach. Notice that $\mathbb{Q}$ is the union of the subgroups $U_n := \frac{1}{n!} \mathbb{Z}$ for $n \geq 0$. We have $U_n \subseteq U_{n+1}$. There is an isomorphism $U_n \cong \mathbb{Z}$ given by multiplication with $n!$. It follows that we have a commutative diagram with vertical isomorphisms $$\begin{array}{c} U_{n-1} & \xrightarrow{\subseteq} & U_n \\ \downarrow && \downarrow \\ \mathbb{Z} & \xrightarrow{n} & \mathbb{Z} \end{array}$$ Hence, $\mathbb{Q}$ is the colimit of the diagram $\mathbb{Z} \xrightarrow{1} \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{3} \dotsc$ This means, by definition, that $\hom(\mathbb{Q},G)$ identifies with the set of compatible homomorphisms $f_n : \mathbb{Z} \to G$. We already know that such a homomorphism corresponds to an element $g_n \in G$, and the compatibility comes down to $n \cdot g_n = g_{n-1}$. This finishes the proof, without any calculations.
Finally let me show how to prove this group presentation without universal properties, but rather with elements: Once you have observed that $\mathbb{Q}$ is generated by the $x_n = \frac{1}{n!}$ and that these elements satisfy $n \cdot x_n = x_{n-1}$, you somehow have to prove that these are the only relations. But how do you want to prove this rigorously? What does it mean at all? Well, by the traditional definition the group $H:=\langle x_0,x_1,\dotsc : n \cdot x_n = x_{n-1} \rangle$ is the quotient of the free group in $x_0,x_1,\dotsc$ modulo the smallest normal subgroup containing $n x_n - x_{n-1}$, i.e. we have $n \cdot x_n = x_{n-1}$ in $H$ (I will denote the residue classes of the $x_n$ also by $x_n$). First of all, observe that $H$ is commutative (so that the additive notation is justified): If $m \leq n$, then the relation $n \cdot (n-1) \cdot \dotsc \cdot (m+1) \cdot x_n = x_m$ shows that $x_n$ commutes with $x_m$. Since the generators of $H$ commute with each other, this means that $H$ is commutative (or abelian). There is a homomorphism $H \to G$ mapping $x_n$ to $x_n=\frac{1}{n!}$, sorry for the abuse of notation (by the universal properties of the free group and of a quotient - you cannot really avoid this!). It is surjective. But how do you prove that it is injective? We have to show that every element in the kernel is already trivial. Well, a priori we only know that every element of $H$ is a possibly long sum of elements of the form $\pm x_n$. But the relation mentioned above proves that actually every element is of the form $z \cdot x_n$ for some $n \geq 0$ and $z \in \mathbb{Z}$: We just take $n$ to be the "maximal" generator occuring in the sum, and then express all the lower ones as multiples of it. Now if some element $z \cdot x_n$ lies in the kernel, this means that $\frac{z}{n!} = 0$ in $\mathbb{Q}$, hence $z=0$ in $\mathbb{Z}$, hence $z \cdot x_n = 0$ in $H$. This proves that $H \cong G$.
I would appreciate any comments which of these three proofs you (the reader) prefer and why.