Proving the powers of a element are all distinct. In a Abstract Algebra textbook of "Dummit and Foote" , There is a Group theory problem at 1.32 . The Problem goes like this :
"If $x$ is an element of finite order $n$ in $G$, prove that the elements $1$ , $x$ , $x$$2$ , $x$$3$ , $x$$4$ , ..... , $x$$(n-1)$ are all distinct.Deduce that  $|x|$ $\le$ $|G|$. "
My Attempt: Lets say there is two numbers $a$ and $b$  and they follow this inequality:
$0$ $\le$ $a$ $\lt$ $b$ $\le$ $(n-1)$
Lets assume that:
$x$$a$ = $x$$b$
$\rightarrow$ $x$$(b-a)$ $=$ $1$
$\rightarrow$ $x$$(b-a)$ $=$ $x$$n$
So , $(b-a) = n$
On the other hand , From the inequality , I can deduce that $(b-a)$ $\lt$ $n$  ,Which is a contradiction .So , $x$$a$ $\neq$ $x$$b$
On the other hand , if $|x|$ $>$ $|G|$ , Then $x$ can generate more distinct elements than the Group even has , which is never possible .So , $|x|$ $\le$ $|G|$.
Confusion: First of all , The question is unclear for me if I choose the order of $x$ to be $0$ or $1$.Because if $|x|$ is $0$ or $1$ , then ,  $x = e$ and all powers of $e$ are the same ($e$).  The problem doesn't tell me what values of $n$ I am restricted to.Can someone tell me what that is?
Also , My attempt seems to be very hand wavy .Is there a way to write the proof with Formal logic in a rigorous manner and where are the major flaws of my proof?. (I am asking this because in the future , I want to learn how to write proper proofs and learn Formal logic).
 A: 
"If x is an element of finite order n in G, prove that the elements 1 , x , x2 , x3 , x4 , ..... , x(n−1) are all distinct. Deduce that |x| ≤ |G|. "

Let's assume that for some two elements $x^a,x^b$ in x's cyclic group (where $0\leq a<b\leq n$) , $x^a=x^b$.

*

*if $a=0$: by the assumption and definition of the first element in a cyclic group, $x^a=x^b=1$. Therefore (by definition of a finite cyclic group), we find our cyclic group is of size $b-1<b\leq n \rightarrow |x|<n$, while by definition, $|x|=n$, a contradiction.

*else ($a>0$): we have a cyclic sub-group $AB=\{x^a,x^{a+1},...,x^{b-1}\}$, and since $0<a<b\leq n$, none of our elements in the cycle contain an identity element, meaning it's an infinite cyclic group. And since $AB \subset X$, we find that x too is of infinite order, a contradiction to the definition that x is of a finite order.

Therefore, the initial assumption, namely that X contains two identical elements in its cyclic group, is false.
Conclusion (1): X's cyclic group elements $1,x,x^2,x^3,x^4,.....,x^{(n−1)}$ are all distinct. $\blacksquare$
From here, assume $|x|>|G|$. Remember that for every elem $a\in x, a\in G$.
By the Pigeonhole principle, x's cyclic group must contain at least one identical pair of values from G, in contradiction to proof (1). Therefore, our initial assumption must be false.
Conclusion (2): $|x|\leq |G| \blacksquare$
A: As you want to prove that distinct powers correspond to distinct elements, you aim to get a contradiction by assuming $\exists a,b, 0\le a<b\le n-1$, such that $x^a=x^b$. This is indeed the case, because this latter would mean that $\exists l(:=b-a), 0<l\le n-1$, such that $x^l=1$, against the minimality of $n$ (by definition of order of an element).
Now, by closure axiom, $\langle x\rangle :=\{1,x,\dots,x^{n-1}\}\subseteq G$, whence $|x|:=|\langle x\rangle|\le |G|$.
