I guess the answer is very trivial, but at the moment I don't get it. Let $p$ prime. Consider the extension $\mathbb{F}_{p^n} | \mathbb{F}_p$ with $n \in \mathbb{N}$ arbitrary. We know that this is a galois extension. But why this extension is separable? Especially for $p|n$ it can't be separable, because there is at least one element who's derivative of his minimal polynomial is zero?
Where is the misstake? Ty for help.
No, there are no elements $\alpha\in\Bbb{F}_{p^n}$ such that the minimal polynomial $m(x)$ of $\alpha$ would have zero derivative. For $m'(x)$ to be zero, we must have the form $$ m(x)=\sum_{i}a_ix^{pi} $$ for some finite set of coefficients $a_i\in\Bbb{F}_p$. By Little Fermat $a_i^p=a_i$ for all $i$, so in fact $$ m(x)=\left(\sum_ia_ix^i\right)^p. $$ But this contradicts the fact that a minimal polynomial is irreducible.