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?

You can always do:
$f(y)=1+y+ \dots +y^{r-1}$ so that $yf(y)=y+y^2+\dots +y^r$ and $yf(y)-f(y)=(y-1)f(y)=y^r-1$
Then put $y=x^d$ with $dr=n$ and obtain $(x^d-1)f(x^d)=x^n-1$ and by construction $f(x^d)$ is a polynomial.
