I am aware that if there is a bifurcating tree with N leaves, then there are (N-1) internal nodes (branching points) with a single root node. How is this relationship proved?
I think the easiest way it to regard this a knock-out competition with $n$ teams, each inner node is a match, and there is one winner. So there are $n-1$ teams to knock out and so $n-1$ nodes.
Or you could do it by induction noting that replacing a pair of leaves and an inner node by a leaf reduces the number of inner nodes and leaves by 1, and that if there is only one leaf then there is no inner node.
I did it myself, the base case
Since the k+1 step relies and follows from the k step the base case can be reached.