Proof: Any $g\in G$ can be written $g=g_1g_2\ldots g_k$ with $g_i\in G_{j_i}, j_i\neq j_{i+1}, g_i\neq e$ since the $G_i$ generate $G$ Thus there is a surjective homomorphism $\phi:\coprod G_i\twoheadrightarrow G$ (since $\coprod G_i$ as the coproduct, has this universal property). We must show $\ker \phi$ is trivial. Choose $g_1g_2\ldots g_k$ as above. Then $g_k(s)\in S_{j_k}$ $g_{k-1}(g_k(s))\in S_{j_{k-1}}$ and so forth, so that $g_1(g_2(\ldots(g_k(s)\ldots)\in S_{j_1}$ But $e(s)=s\notin S_{j_1}$ Thus $\phi(g_1g_2\ldots g_k) \neq e$ and $\phi$ is injective.