I think you can use induction to the number of vertices. Let G be a graph with only 1 vertex. To satisfy the condition "e is not less than v", it must have a loop. So, G has a cycle.
Now suppose that the statement is true for all graphs with number of vertices at most k. Consider a graph G with (k+1) vertices and suppose that |E(G)| is not less than |V(G)|. If G has no cycles, than it must have a "leaf", v_0. (A leaf means a vertex of a graph of degree 1.)
Now consider the graph H we get by removing the vertex v_0 from G. Then H is a graph with k=|V(G)|-1 vertices and |E(G)|-1 edges. So, the number of edges of H is not less than the number of vertices of H. By the induction hypothesis, H must contain a cycle. However, since H is a subgraph of an acyclic graph G, H cannot contain a cycle. This is a contradiction.
Now we see that the statement holds for all graphs with finite number of vertices.