Graph Theory Mini Proof Check (Leaves) I'd just like to prove the existence that at least one leaf node exists in any spanning tree (I'm using this fact in another proof and want to make sure I'm being as rigorous as possible).  Here's my attempt:
1) Suppose for contradiction that a leaf node does not exist in a tree, H.
2) This would imply every node in the tree has a degree of 2 or higher.
3) If there are at least two edges incident to every node, then there are at least two unique (undirected) paths between any two nodes in the tree.
4) Let v and w be two nodes in the tree, and node v is incident to two edges, e and f.
5) By using the definition of a circuit, edges e and f are both part of a circuit if and only if after removing an either e or f, there still exists a simple path between nodes v and w.
6) There will in fact exist a simple path between v and w after e is removed (or vice versa if f is removed), with the path containing edge f (or e if f is removed).  This implies that edge e (and f) are part of a circuit.
7) If H is a tree, there cannot be circuit.  This is a contradiction to our original assumption, and, as a result, H cannot be a tree with all nodes having two or more edges.
I know there are several ways to prove this, but I wanted to know if this is an adequate way.  I'm pretty new to proofs (and graph theory), so any feedback would be great. 
 A: As it stands, your argument is seriously incomplete: your third assertion requires justification, and none is given. Starting with the assumption that each vertex has degree at least $2$ it’s at least as easy to prove directly there’s a circuit as it is to prove your third assertion. (Start at any vertex. Walk from vertex to vertex, always going to an unvisited vertex if possible. If every vertex has degree at least $2$, you never reach a dead end, and the tree has only a finite number of vertices, so you must reach a point at which every edge at your current vertex leads to a vertex that you’ve already visited. At that point you have a circuit.)
If you’ve already proved that a tree always has exactly one more vertex than it has edges, you can prove that every tree has at least one leaf by observing that if every vertex of a graph has degree at least $2$, and the graph has $n$ vertices, then the sum of the degrees of the vertices is at least $2n$, and by the handshaking lemma the graph must then have at least $n$ edges and cannot be a tree.
