 Solution

A primitive root of unity ${\xi }_n$ satisfies ${\xi }_n^n=1$ and ${\xi }_n^j\ne 1$ 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 $a\cdot 1$ and $b\cdot 1$ would have product zero. If this prime $p$ divides $n$, put $m=n/p$. In characteristic $p$, the binomial coefficients $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ 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 $\boxed{\mathrm{char}K\nmid n}$.