Let $K$ be a finite and normal extension of $F$ and suppose that the extension $K/F$ has no proper intermediate extension. Show that the degree $[K:F]$ is a prime number.
Do we consider the Galois group $\text{Gal}(K/F)$ and that $[K:F]=|\text{Gal}(K/F)|$ ?
Since the extension $K/F$ has no proper intermediate extension do we get that the Galois group has no proper subgroup? And that would mean that there is no $n$ such that $n\mid |\text{Gal}(K/F)|$ with $n\neq 1$ an $n\neq |\text{Gal}(K/F)|$ ?
As the commenters pointed out, the possibility of inseparability adds a few extra twists.
If the extension $K/F$ is also assumed to be separable, then it is Galois. Furthermore, lack of intermediate fields is then equivalent to the Galois group not having non-trivial subgroups. That is easily seen to be equivalent to the Galois group being cyclic of prime order proving the claim in that case.
If the extension $K/F$ is not known to be separable, then we can proceed as follows. Let $\alpha\in K\setminus F$ be arbitrary. Because there are no intermediate fields, you can deduce that $K=F(\alpha)$. Let $m(x)\in F[x]$ be the minimal polynomial of $\alpha$ over $F$. If $m(x)$ is separable, then so is $K/F$ and we are back in the first case. Therefore we can deduce that $m(x)$ has multiple zeros, and the usual argument then shows that we must be in positive characteristic, say $p$, and that $m'(x)=0$ implying that $$ m(x)=x^{np}+a_{n-1}x^{(n-1)p}+\cdots+a_1x^p+a_0. $$ In other words $m(x)=f(x^p)$ for the polynomial $f(x)=x^n+\sum_{i=0}^{n-1}a_ix^i\in F[x]$.
Here obviously $f(\alpha^p)=m(\alpha)=0$. Therefore the minimal polynomial of $\alpha^p$ is a factor of $f(x)$. But $f(x)$ is necessarily irreducible for if $f(x)=g(x)h(x)$ non-trivially then $m(x)=g(x^p)h(x^p)$ in violation of irreducibility of $m(x)$. So $f(x)$ must be the minimal polynomial of $\alpha^p$.
Because $\deg f(x)<\deg m(x)=[K:F]$ we can deduce that $F(\alpha^p)$ is a proper subfield of $K$. The lack of intermediate fields means that $F(\alpha^p)=F$. In other words, $\alpha^p$ is an element of $F$. But this implies that the minimal polynomial of $\alpha^p$ over $F$ is the linear polynomial $f(x)=x-\alpha^p$. Therefore $$m(x)=f(x^p)=x^p-\alpha^p.$$ Hence $[K:F]=\deg m(x)=p$. QED.
