= Solution
A <primitive root of unity> $\xi_n$ satisfies $\xi_n^n=1$ and $\xi_n^j\ne1$ for $1\le j<n$, equivalently it has multiplicative order $n$. A positive <characteristic of a field> must be prime: if its least positive value $p$ factored as $ab$ with $1<a,b<p$, the two nonzero field elements $a\cdot1$ and $b\cdot1$ would have product zero. If this prime $p$ divides $n$, put $m=n/p$. In characteristic $p$, the binomial coefficients $\binom pj$ for $0<j<p$ vanish because their integer numerators contain $p$ and their denominators do not. Thus $(X-1)^p=X^p-1$. It follows that $(\xi_n^m-1)^p=\xi_n^n-1=0$. A field has no nonzero nilpotent element, so $\xi_n^m=1$, contradicting primitivity. Consequently $\boxed{\operatorname{char}K\nmid n}$.
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