Graph Theory: Tree ($n\geq 4$ nodes) with no vertex degree $2$. Prove there is at least one vertex w/ 2 or 3 leaves as neigbors T: Trees with no vertex of degree 2 have more leaves than internal nodes
So far I have (proof by contradiction).
Consider the opposite. That all nodes have only 1 leaf as a neighbor.
Take some vertex, $v_i $. It must be connected to 2 other vertices which must be internal nodes (deg(3)).
However this would mean there could exist a tree with more internal nodes (2) than leaves which would contradict T.
Hence the original claim is false.
Hence there is at least one vertex with 2 or 3 leaves.
Feel like this isn't as watertight as I would like.

One possible proof by contradiction (welcome comments)
Assume each node has max. 1 neighbor with is a leaf.
--> Every inner node has at least 2 neighbors which are inner nodes.
So now choose some inner node v.
Visit a neighboring inner node v', then another v'' != v, etc..
After a finite number of steps we end up at an inner node already visited {I believe this is because there is always a way in and out of a node given min. 2 in-degree to any inner node}, i.e. there is a cycle.
This contradicts the original assumption that we have a Tree (no cycles).
--> Hence each node must have 2 or 3 neighbors which are leaves. 
