Let G be a group. Suppose that x ∈ G has order n ∈ N. Prove that x^n = 1. I feel like the maths at university is very abstract and sometimes I get lost into the way things are written.(as an example a simple binary operation can be generally noted as a*b or just ab, depending on the writer's idea) 
I guess that can be only fixed by reading lots of proofs?
I think this "abstract" idea applies on the above question from the title as well as I am stuck.
Something I can think about is:
If we have got a group of integers with group operation being multiplication.
I let x=4 then x^n is not 1. So the proof doesn't make sense in this situation for me.
solution:
We have  < x > is a set of size n. The non-negative powers of x are
1, x, x^2, . . . which are not all different because < x > is finite. Let the first repetition be x^m = x^q with 0 ≤ m < q. If m > 0 we can premultiply by x^(−1)
to get an earlier repetition (which is impossible), so m = 0. Thus 1, x, x^2
, . . . are different until a 1 appears as x^q. Therefore q = n. The order of x is n and also x^n = 1.
 A: It seems that the definition of order you're using is

$x$ has order $n$ if (and only if) $\langle x\rangle=\{x^k:k\in\mathbb{Z}\}$ has $n$ elements.

Note that $\langle x\rangle$ is the image of a group homomorphism $\varphi\colon\mathbb{Z}\to G$, $\varphi(a)=x^a$. The kernel of $\varphi$ is a subgroup of $\mathbb{Z}$, so it is of the form $k\mathbb{Z}$. By the homomorphism theorem,
$$
\mathbb{Z}/k\mathbb{Z}\cong\langle x\rangle
$$
so $k=n$ by counting elements. Hence $n\in\ker\varphi$, which precisely means that $x^n=1$.
Without homomorphisms, your attempt is quite good, but you're too fast in the conclusion.
You prove correctly that $m=0$, so $x^q=1$. But this only proves that $q\le n$, because the elements $x^0,x^1,\dots,x^{q-1}$ are all distinct.
On the other hand, given any integer $p$, you can write $p=qp'+r$, with $0\le r<q$ and
$$
x^p=(x^q)^{p'}x^r=x^r
$$
so the elements in $\langle x\rangle$ are all of the form $x^r$, with $0\le r<q$. Since there are $n$ elements by assumption, it follows that $n\le q$.
