Make sure the edges of the tree are ordered to point away from the root $r$. So, for all edges $(u, v) \in E$, make sure that $d(r, v) = d(r, u) + 1$.

Create a function $terminal$ which maps every outward arc to its target. $terminal: E \rightarrow V$, $terminal((u, v)) = v$.

This map gives us an almost bijection from edges to all vertices other than the root. So we have that $|E| + 1 = |V|$. Each of the edges cover one non-root vertex, and we then $+1$ to count the root node.

I found this much more intuitive than the inductive argument. I feel like I should attempt to "parallelize" inductive arguments so you can see the entire counting "at once".