If L is finite extension of characteristic p field K with $gcd([L:K],p)=1$, show L is separable. In Lang's Algebra, I had hard times with this question:

Let $K$ be field with characteristic $p$ (a prime). Let $L|K$ be a finite extension of $K$, and suppose $gcd([L:K],p)=1$. Show $L$ is separable over $K$.

In order to have $L$ separable over $K$, one needs to have $K(\alpha)|K$ separable for each $\alpha \in L$. 
We can say $L=K(\alpha_1,...,\alpha_m)$. It seems like I need to use the tower relation of these extensions and get some information about their degrees that will be relevant to the gcd thing. But I couldn't do these. A hint is welcomed.
 A: The extension $K(\alpha)/K$ is separable when the minimal polynomial of $\alpha$ over $K$ has no repeated roots.
An irreducible polynomial $f\in K[x]$ has repeated roots if and only if $f(x)=g(x^p)$ for some $g\in K[x]$.
Do you see how to proceed from here?
A: For every $\alpha \in L$, the minimum polynomial $f_\alpha(x)$ has a degree dividing the degree of the extension. As $[L : K]$ is relatively prime to $p$, we get that the degree of $f_\alpha(x)$ must also be relatively prime to $p$.
Then, as $f_\alpha(x)$ is irreducible, we can write
$$ f_\alpha (x) = a_q x^q + \cdots + a_0 $$.
Then, its derivative is
$$ f'_\alpha (x) = q \cdot a_q x^{q-1} + \cdots + a_1 $$.
As $F$ is an integral domain, $q \cdot a_q \neq 0$ for non-zero $q$ and $a_q$. This shows that $f'_\alpha (x) \neq 0$. Hence, $f_\alpha(x)$ is separable.
Reference: Separable Extension given prime characteristic. The only thing left is that, in the extension $F/K$, why the degree of any minimum polynomial divides the degree of the extension. Would appreciate any suggestions!
Hope it can help!
