k-th power of all roots of unity over $\mathbb{K}$

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Let $n \geq 2$ and let $\omega_1,...\omega_n$ be all the $n$th roots of unity, not necessarily distinct over an arbitrary field $\mathbb{K}$. Prove that

$\omega_1 ^k + \omega_2^k+...+\omega_n^k = n$, if $k = 0$ and

$\omega_1 ^k + \omega_2^k+...+\omega_n^k= 0$, if $k = 1,...,n-1$

I think I should somehow make use of the the fact that

$1+ \omega + \omega^2+...+\omega^{n-1} = 0$, if $\omega\neq1$ and

$1+ \omega + \omega^2+...+\omega^{n-1} = n$, if $\omega=1$

But I don't see how I can use it since we have power $k$. Can anyone give me a hint? thanks

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$$w_k=e^{2i\pi k/n},$$ then use the summation of a G.P to get $$S=\sum_{j=0}^{n-1} w^k_j=\sum_{j=0}^{n-1} e^{2i\pi jk/n}= \frac{ e^{2i \pi k}-1}{e^{2i\pi k/n}-1}=0, ~if~k\ne pn, p \in I$$ If $k=pn$, then $$S=\sum_{j=0}^{n-1} e^{2i\pi j p}=n,$$ as $e^{2i \pi j p}=1.$