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?

Hint from Dummit and Foote: 
Suppose $n= kd+r$, where $0 \leq r < d$ we have 
$$x^n -1 = x^{kd+r} - x^r + x^r -1 = x^r (x^{kd}-1) + (x^r -1)$$
then 
can you conclude $d|n$ if and only if $x^d-1 | x^n-1$.
