$G$ solvable $\implies$ composition factors of $G$ are of prime order.
Solution 1:
Hint: Use induction on the order of $G$:
If $G_{s-1}$ is a proper normal subgroup of $G$ such that $G/G_{s-1}$ is abelian, then there are two cases:
- Either $1 < G_{s-1}$; then by induction, the composition factors of $G_{s-1}$ and $G/G_{s-1}$ have prime orders.
- Or $1 = G_{s-1}$ in which case $G$ is itself abelian.
Can you figure out how the claim follows in each of the two cases?