Let us have a field $K\supseteq E$ and $G$ be its group of automorphisms over $E$. Let the fixed field of $G$ be $K^G$. I would like to show that $K$ is separable over $K^G$.
I know that for algebraic extensions normal and separable iff Galois and $K$ is Galois over $K^G$. However, I want to show it a more direct way. I was wondering if there is someway to show that if some irreducible polynomial $f$ was not separable, than some something in its splitting field (which must be contained $K$) that is not in $K^G$ must be also fixed by $G$ leading to a contradiction. However, I cannot get this to work out. This would make sense if $f$ had only a single root repeated, but that is not true in general. (If need be we can also include the assumption that $K$ is normal of $E$, but I do not know if this is necessary)
Any help or direction is greatly appreciated.