If $N$ is the normal subgroup of $A\ast B$ generated by $A$, then $(A\ast B)/N\cong B$

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Suppose that $A\ast B$ is the free product of the groups $A$ and $B$, and $N$ is the normal subgroup of $A\ast B$ generated by $A$. I wish to prove that $(A\ast B)/N\cong B$.

Suppose that $h:A\ast B \rightarrow B$ is the function taking $m_1m_2...m_j$ to $m_{i_1}\ast m_{i_2}\ast...\ast m_{i_s}$ where {$i_t| t=1,2,...,s$} is the set of all $i$ such that $m_i\in B$. I could show that this function is an epimorphism. I also could show that $N$ is contained in $Ker (h)$. If it can be proved that $N=Ker(h)$ the problem is solved. How can I proceed to prove this last part?