Ok, I finally got through it.
Let $K$ be the splitting field of $f$ over $F$. $K/F$ is Galois with degree $p$.
If $K$ lies in a radical extension $L$ of $F$. Then we have
$$
F=F_0\subset F_1\subset F_2\ldots\subset F_r=L
$$
where $F_i=F_{i-1}(\alpha_i)$ and $\alpha_i^{n_i}\in F_{i-i}$. We may assume that $\alpha_i\notin F_{i-1}$ and $n_i$ are all primes.
Let $K_i$ be $K(\alpha_1,\ldots\alpha_i)$, then $F_i \subset K_i$.
By induction, we can prove that $K_i/F_i$ is Galois with degree $p$ as follows.
First, $K_0/F_0$ is Galois with degree $p$.
We assume $K_{i-1}/F_{i-1}$ is Galois with degree $p$. $K_i=K_{i-1}(\alpha_i), F_i=F_{i-1}(\alpha_i)$. If $\alpha_i\in K_{i-1}$, then $F_i=F_{i-1}(\alpha_i)=K_i=K_{i-1}(\alpha_i)=K_{i-1}$ and $n_i=p$, since $[K_{i-1}:F_{i-1}]=p$. Because $\alpha_i\notin F_{i-1}$, $g=(t-\alpha_i)^p=t^p-\alpha_i^p$ is irreducible over $F_{i-1}$. Then the minimal polynomial of $\alpha_i$ over $F_{i-1}$ is $g$. However, $K_{i-1}/F_{i-1}$ is Galois, so $\alpha_i$ is separable, but $g=(t-\alpha_i)^p$, which shows that $\alpha_i$ is not separable.
This contradiction shows that $\alpha_i\notin K_{i-1}$. Note that $\alpha_i^{n_i}\in F_{i-1}\subset K_{i-1}$ and all $n_i$th roots of unity is in $F$. Then we have $g=t^{n_i}-\alpha_i^{n_i}$ is irreducible over $K_{i-1}$ and $[K_i:K_{i-1}]=n_i$. Then we can conclude that $K_i/F_i$ is Galois with degree $p$.
By induction, $K_i/F_i$ is Galois with degree $p$ for all $i$.
On the other hand, $K_r=F_r=L$, so $K_r/F_r$ is of degree 1, which leads to a contradiction.
Best Answer
Definition 1 Let $K$ be a field of characteristic $0$. Let $p$ be a prime number. Let $\alpha$ be an element of an algebraic closure of $K$ such that $\alpha^p \in K$ and $X^p - \alpha^p$ is irreducible over $K$. Then we call $K(\alpha)/K$ a simple prime radical extension.
Definition 2 Let $K = K_0 \subset K_1 \subset \cdots \subset K_n = L$ be a tower of fields. If $K_i/K_{i-1}$ is a simple prime radical extension for $i =1, \dots, n$, we call $L/K$ is a prime radical extension.
Definition 3 Let $L/K$ be a field extension. If there exists a prime radical extension $E/K$ such that $L$ is a subfield of $E$, we call $L/K$ a prime radically solvable extension.
Lemma 1 Let $K \subset M \subset L$ be a tower of fields. Suppose $M/K$ and $L/M$ are prime radically solvable extensions. Then $L/K$ is also a prime radically solvable extension.
Proof: This is proved in Lemma 10 of my answer to this question.
Lemma 2 Let $L/K$ be a finite Galois extension of prime degree $p$. Suppose $char(K) = 0$ and $K$ has a $p$-th primitive root of unity. Then $L/K$ is a simple prime radical extension.
Proof: By this question, there is an element $\alpha$ of $L$ such that $L = K(\alpha)$ and $\alpha^p$ is an element of $K$. Since $p = [L\colon K]$, $X^p - \alpha^p$ is irreducible over $K$. Hence $L/K$ is a simple prime radical extension. QED
Lemma 3 Let $L/K$ be a finite Galois extension of prime degree $p$. Suppose $char(K) = 0$. Then $L/K$ is a prime radically solvable extension.
Proof: Let $\Omega$ be an algebraic closure of $L$. Let $\zeta$ be a primitive $p$-th root of unity in $\Omega$. Then $L(\zeta)/K(\zeta)$ is a Galois extension and $Gal(L(\zeta)/K(\zeta))$ is isomorphic to a subgroup of $Gal(L/K)$. Hence it is a cyclic group of order $p$ or $1$. By Lemma 2, $L(\zeta)/K(\zeta)$ is a prime radically solvable extension. On the other hand, by Lemma 6 of my answer to this question, $K(\zeta)/K$ is a prime radically solvable extension. Hence, by Lemma 1, $L(\zeta)/K$ is a prime radically solvable extension. Hence $L/K$ is a prime radically solvable extension. QED
Theorem Let $K$ be a field of characteristic $0$. Let $L/K$ be a finite Galois extension. Suppose $G = Gal(L/K)$ is sovable. Then $L/K$ is a prime radically solvable extension.
Proof: There exists a tower of subgroups of $G$: $G = G_n \supset G_{n-1} \supset \cdots \supset G_1 \supset G_0 = 1$, where each $G_i/G_{i-1}$ is a cyclic group of prime degree. Hence there exists a tower of fields: $K_0 = K \subset K_1 \subset \cdots \subset K_n = L$, where each $K_i/K_{i-1}$ is a cyclic Galois extension of prime degree. By Lemma 3, $K_i/K_{i-1}$ is a prime radically solvable extension. Hence, by Lemma 1, $L/K$ is a prime radically solvable extension. QED