I am working on the following problem in my Topology class. While I do have an idea of how this proof should work I am having a hard time making into a nice proof. Here is the question and my attempt
A finite graph is called a tree if it is nonempty, connected and contains no cycles. Use the fact that a finite graph with no cycles contains a vertex with $\leq 1$ edges.
Since our graph is nonempty there exists at least one vertex. Let this vertex be a point in a tree such that it has $\leq1$. Then we know this vertex has either 0 edges extending from it or 1 edge extending from it.
case 1. this vertex has 0 edges extending from it. This gives us the single vertex so we have $n=0$ edges and $n+1$ vertices.
case 2. this vertex has one edge extending from it. In this case the edge must end a point that is not the original vertex (no cycles). Thus, we have a single edge connected by 2 vertices. We may also see that for any edge added to a vertex on this tree we must have a new endpoint (again there are no cycles). So, for every edge added there is also a vertex added. Continue adding edges and vertices to obtain our tree. Thus, for $n$ edges we have $n+1$ vertices.