Problem related to $n^{th}$-root of unity

Question

Let $n>1$ be an integer, $k$ be an integer such that gcd(k,n)=1 and $w\neq 1$ be an $n^{\text{th}}$ root of unity. Show that, $$1+w^k+w^{2k}+\ldots+w^{(n-1)k}=0$$ I tried this problem in this way let $w=e^{\frac{2\pi ip}{n}}$, where $1\leq p\leq n-1$. If $w^{k}\not=1$ then we have done because\ $$1+w^k+w^{2k}+\ldots+w^{(n-1)k}=\frac{1-w^{nk}}{1-w^{k}}.$$ Now using the condition $gcd(n,k)$ how to show that $w^{k}\not=1.$

Answer 1

If $w^k=1$ and $w^n=1$ and $\gcd(n,k)=1$ impliy that $w=1$ by Bézout's identity.

Indeed, if $ak+bn=1, \enspace(a,bv\in\mathbf Z)$, then $$w=w^{ak+bn}=(w^k)^a\cdot(w^n)^b=1^a\cdot1^b=1.$$