Understanding proof that if x has finite order, the n powers repeat

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So we want to prove this theorem from Group Theory:

Let $x$ be an element of a group $(G, \circ)$. If $x$ has finite order $n$, then the $n$ powers

$$e, x, x^2, ..., x^{n-1}$$

are distinct, and these elements repeat indefinitely every $n$ powers in the list of consecutive powers of $x$.

Then my book uses a proof by contradiction to show that the powers are unique. After that they continue to show that the powers repeat every $n$ elements:

Consider any power of $x$ of the form $x^{kn}$. That is, any power where the exponent is an integer multiple of the order $n$ of $x$. We have:

$$x^{kn} = (x^n)^k = e^k = e$$

So in the list of consecutive powers of $x$, the element $e$ is repeated every $n$ elements.

Because each element in the list is obtained from the previous element by composing it with $x$, it follows that all the elements in the list repeat every $n$ elements.

I'm grappling with the sentence in bold. I agree that the proof shows that $e$ repeats every $n$ powers.

I also agree that each element in the list of consecutive powers is obtained from the previous element by composing it with $x$.

But somehow I can't get why this implies that all elements repeat.

(I know it's true, I just find it hard to accept the proof in the book shows it properly)

Can someone add some insight why the proof given is sufficient?

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The bold statement is just saying, since $x^{kn} = e$, composing with $x$ on both sides shows how all elements repeat:

$$\begin{align*} x^{kn} &= e\\ x^{kn+1} &= x\\ x^{kn+2} &= x^2\\ &\vdots\\ x^{kn+i} &= x^i \end{align*}$$

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$$x^r= e(x^r)=x^nx^r=x^{n+r}$$ Therefore each term $x^r$ repeats.