 Solution

Let $\alpha :G\to G$ be surjective and suppose that $1\ne g\in \ker \alpha$. By residual finiteness, choose $q:G\to Q$ with $Q$ finite and $q(g)\ne 1$. The preceding part makes the sequencerepeat, so $q{\alpha }^i=q{\alpha }^j$ for some $i<j$. Surjectivity of ${\alpha }^i$ permits cancellation on the right and gives $q=q{\alpha }^{j-i}$. But $g\in \ker {\alpha }^{j-i}$, which would imply $q(g)=1$, a contradiction. Thus $\alpha$ is injective. Every finitely generated residually finite group is therefore a Hopfian group.