Simple pathes in an acyclic graph
Since the graph is acyclic, we know that it is a forest - which means that, there is at most $1$ simple path between two nodes. Now, simply root the graph at one vertex of the edge, remove the edge, and run a dynamic programming algorithm to find the number of vertices in the subgraph, call it $v_a$. Similarly, do the same for the other vertex of the edge, call it $v_b$. The answer is $v_a*v_b$, since the simple path between every $2$ nodes is a different simple path.

Proof: Clearly every simple path we consider passses through the chosen edge. The only other simple paths are the ones where both the nodes are chosen in the same subgraph, which clearly does not pass through this edge.