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Now for the inductive case, fix \(k \ge 1\) and assume that all trees with \(v=k\) vertices have exactly \(e=k-1\) edges. Now consider an arbitrary tree \(T\) with \(v = k+1\) vertices. By Proposition 4.2.3, \(T\) has a vertex \(v_0\) of degree one. Let \(T'\) be the tree resulting from removing \(v_0\) from \(T\) (together with its incident edge). Since we removed a leaf, \(T'\) is still a tree (the unique paths between pairs of vertices in \(T'\) are the same as the unique paths between them in \(T\)).