Let $G$ be finite group, and assume that every maximal subgroup of $G$ has prime index.
Show that $G$ is solvable.
I tried to prove it myself, but I'm afraid there may be mistakes in my proof.
$\textit{Proof:}$
To begin with, let's use the fact that
If all maximal subgroups of $G$ have a prime index, then there is a normal Sylow $p$-subgroup of $G$, where $p$ is the largest prime divisor of $|G|$.
I will leave this without proof, since this statement is easy to prove.
We prove this statement by using induction on the number of distinct prime divisors of order $G$. Let $n$ be the number of distinct prime divisors of $|G|$.
If $n = 1$, then this is true (for example, because any $p$-group is nilpotent).
Now let $|G| = p_1^{\alpha_1} \ldots p_n^{\alpha_n}.$ Without limiting generality, we can assume that $p_1 < p_2 < \ldots < p_n$.
Then, using the above fact, there exists normal $S\in{\rm Syl}_{p_n}(G)$. Then $(|S|, |G:S|) = 1$ and since $S$ is normal, by the Schur-Zassenhaus theorem, we can conclude that $S$ has a complement $H$ in $G$.
$S$ is solvable, and $|H|$ has fewer distinct prime divisors than $|G|$ $\Rightarrow$ by induction assumption $H$ is solvable, but at the same time $H\cong G/S\Rightarrow S$ and $G/S$ are solvable $\Rightarrow G$ is solvable. $\blacksquare$