<s>[INST] if $d\mid n$ then $x^d-1\mid x^n-1$ proof How would you show that if $d\mid n$ then $x^d-1\mid x^n-1$ ?
My attempt :

$dq=n$ for some $q$. $$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$ in fact, $$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$
By multiplying both sides of $(1)$ by $(x-1)$ we get that $1-x^d\mid 1-x^n$ which is the final result

Is this an ok proof?
 [/INST] You have:
$$d\mid n \Rightarrow n=ad$$
Then:
$$x^n-1=x^{ad}-1=(x^d)^a-1$$
Setting $x^d=y$ (just for simplifying the process) we have:
$$y^a-1=(y-1)(y^{a-1}+\dots+y+1)=(x^d-1)((x^d)^{a-1}+\dots + x^d+1)$$
In other words we showed that:
$$x^n-1=   (x^d-1)((x^d)^{a-1}+\dots + x^d+1) $$
Which obviously implies that:
$$x^d-1 \mid x^n-1$$
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