Every minimal normal subgroup of a finite solvable group is elementary abelian

First, note that $H$ is abelian. Indeed, we know that $G$ is solvable, and since $H\unlhd G$ this implies that $H$ is solvable. Now, $[H,H]$ is a characteristic subgroup of $H$, and thus normal in $G$, and so by assumption this implies that $H=[H,H]$ or $\{1\}=[H,H]$. If the former happened then $H$ would not be solvable (why?) and so the latter must happen, which says precisely that $H$ is abelian.

Let's show next that $|H|$ is divisible by only one prime. Let $p\mid |H|$, then $H$ has a $p$-Sylow subgroup $S$. Since $H$ is normal in $G$ and $S$ is characteristic in $H$ (because it's the unique Sylow subgroup since $H$ is abelian) we must have that $S$ is normal in $G$, by assumption this implies that $H=S$ so that $|H|=p^n$ for some $n$.

Finally, since $H$ is an abelian $p$-group we know that $pH$ is a proper (by Cauchy's theorem) of $H$ which is necessarily normal in $G$ (since it's characteristic in $H$), and so $pH=\{1\}$. This implies that $H=\mathbb{F}_p^n$ as desired.