<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] Let $\alpha$ be a solution to the equation $x^d = 1$.  We then have that $\alpha^d$ = 1.  Since $d|n$ we can write $n = d\cdot k$ for some integer $k$.  Thus $$1 = 1^k = \left( \alpha^d \right)^k = \alpha^{d\cdot k} = \alpha^n$$
This shows that $\alpha$ is a solution to $x^n = 1$.  
This shows that $x^d-1|x^n-1$.    
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