How to prove that a connected graph with $|V| -1= |E|$ is a tree? I could neither show myself nor find a proof of the following result:  if $G=(V,E)$ is a connected graph with $|E|=|V|-1$ then $G$ is a tree.
Could somebody please provide an argument to establish this.
 A: An empty graph on $n$ vertices has $n$ connected components, Suppose you have a graph and add an edge, then the number of connected components is reduced by at most one ( since this edge touches at most two connected components). Therefore a connected graph on $n$ vertices has at least $n-1$ edges).
Suppose a connected graph on  $n$ vertices has $n-1$ edges, we must prove no cycle exists, suppose it does, if we remove an edge from the cycle we get a connected graph (because if we have a path that uses this edge, instead of using the edge we can go around the remaining part of the cycle). This is a contradiction, since the graph would have $n-2$ edges, and would hence not be connected.
A: Suppose $|V| \ge 2$. Recall that the sum of the degrees of all vertices is $2|E|$. Since The graph is connected there cannot be a vertex of degree $0$. 
Thus as $2|V|> 2|E|$ there is a vertex of degree one. Removing that vertex and the adjacent edge does not change that the graph is connected. And if the graph after removal is a tree, so is the graph itself. 
Using this starting idea you can set up a proof by induction.
A: Suppose the graph contains cycles, then we can remove an edge without removing a vertex and the graph is still connected. After removing, it will has $n$ vertices and $n - 2$ edges. Continue to do so until we reach a tree (because there is no cycle) with $n$ vertices and $n - k$ edges $(k > 1)$. However it contradicts with the fact that a tree with $n$ vertices must has $n - 1$ edges.
Therefore the graph must not contain cycles, and then it is a tree by definition.
