I want build a separable extension $E/F$. Suppose that $E/F$ is a finite divisible field extension. I want to prove that $E/F$ is separable in this method:
we know that if $Char(F)=0$, then $E/F$ is separable. So suppose that we are in case $Char(F)=p>0$ for a prime $p$ and $F$ is indivisible. We know that $F^*=N_{E/F}(E^*)\times C_m$ for $1\neq m|[E:F]=n$ and From ${F^*}^n\subseteq N_{E/F}(E^*)$, every element of $C_m$ is a root of $x^n-1$. So if I can say $p\nmid m$, I can prove that ${F^*}^p=F^*$, and from it I can prove that $F$ is perfect and so $E/F$ is separable.
So my question is why $p\nmid m$?
As $C_m$ is a cyclic group of order $m$, so we have $m$th root of unity in $F$ like $a$. If $p=Char(F)|m$, then there exists natural number $r$ such that $rp=m$ we have :
$0=a^m-1=(a^r)^p-1=a^r-1$, so the order of $a$ can not be $m$ and it is contradiction. Therefore $p\nmid m$.