The proof uses character theory. We outline the key steps.

Step 1 (Burnside's lemma on conjugacy classes): If $\chi$ is an irreducible character of $G$ with $\gcd(\chi(1), |C|) = 1$ for some conjugacy class $C$, and $g \in C$ with $g \neq e$, then either $\chi(g) = 0$ or $|\chi(g)| = \chi(1)$ (in which case $g \in Z(\rho(G))$ for the corresponding representation $\rho$).

Step 2: If $G$ is a non-abelian simple group of order $p^a q^b$, consider a conjugacy class $C$ with $|C| = p^k$ (such a class exists since not all classes have size divisible by $q$). Every irreducible character $\chi$ with $p \nmid \chi(1)$ satisfies $\gcd(\chi(1), |C|) = 1$, so by Step 1, either $\chi(g) = 0$ or $g \in \ker \rho$. Since $G$ is simple, $\ker \rho = \{e\}$ or $G$.

Step 3: Using the column orthogonality relation $\sum_\chi |\chi(g)|^2 = |C_G(g)|$ and the fact from Step 2, derive that $g$ acts trivially in too many representations, leading to $g = e$ -- a contradiction.

Therefore no non-abelian simple group has order $p^a q^b$, meaning any group of order $p^a q^b$ has a nontrivial normal subgroup. By induction on $|G|$, both the normal subgroup and the quotient are solvable, hence $G$ is solvable. $\blacksquare$