I am interested in an intuitive proof for "any binary tree with $n$ nodes has $n-1$ edges", that goes beyond proof by strong induction.
You can see a (binary) tree as a directed graph: suppose the root is the "lowest" node and the leaves are the "highest" ones, then say that all the edges are oriented upwards. Then, every node that is not the root will have exactly one edge entering in it, and every edge will be pointing at exactly one node. This means that if you have $n$ nodes, you have to have $n-1$ edges (one per node with the exception of the root).
This is essentially a different perspective on OmG's answer.
The technical rendition is: A tree is contractible (i.e. homotopic to a point). Since the Euler characteristic (the number of vertices minus the number of edges) is a homotopy invariant, any tree has the same Euler characteristic as a point which is $1$ (one vertex minus zero edges).
The non-technical rendition is: If we collapse a leaf into its parent, we lose one edge and one node, so the difference between the number of edges and the number of vertices remains the same. Clearly, we can just keep doing this until we're left with only the root node for which the difference between vertices and edges is $1$ (one vertex and zero edges).
More generally, collapsing two distinct vertices along an edge that joins them is a homotopy, so for any graph at all we could collapse any distinct pair of vertices joined by an edge without changing the Euler characteristic. That means in the above we didn't need to work from leaves to root; we could have collapsed any edge, though that would have led to intermediates that aren't binary trees. It also means that since a loop (one vertex and one edge from that vertex to itself) has Euler characteristic $0$, then any cycle has Euler characteristic $0$.