There are m distinct pᵏmth roots of unity over a field of characteristic p, for p a prime not dividing m

Let $n = p^km$, where $p$ is a prime not dividing $m$, and let $F$ be a field of characteristic $p$. Prove that there are $m$ distinct $n$th roots of unity over $F$.

Note that $x^n-1 = x^{p^km}-1$ $= (x^m)^{p^k}-1$ $= (x^m-1)^{p^k}$. That is, the distinct $n$th roots of unity over $F$ are precisely the distinct roots of $x^m-1$ over $F$.

If $m=1$, then certainly there is only 1 root of $x-1$.

Suppose $m > 1$. Now $D(x^m-1) = mx^{m-1}$, which has only the root 0 with multiplicity $m-1$. Clearly 0 is not a root of $x^m-1$, so that $x^m-1$ and its derivative are relatively prime, and thus $x^m-1$ is separable. Hence there are $m$ distinct $m$th roots of unity over $F$, and so $m$ distinct $n$th roots of unity over $F$.