Any graph on $n$ vertices with $k$ automorphisms – ways its vertices can be mapped onto themselves to preserve the edges – has $\frac{n!}k$ labellings. Applying this to the trees on seven vertices (the image below taken from Peter Steinbach's Field Guide to Simple Graphs, available on the OEIS under the links at A000055):

the number of labelled trees isomorphic to $G$, which is given by $\frac{6!}{|\mathrm{Aut}(G)|}$ by the Orbit-Stabiliser Theorem. We see that $$6+120+360+90+360+360=1296=6^4$$ which matches Cayley's Tree Formula, so no tree is missing.

How must a vertex attach to a tree so that the resulting graph is also a tree? How many leaves can a tree have? Draw all non-isomorphic trees with at most 6 vertices?

It is well known that the number of labelled trees on $n$ vertices is equal to $n^{n-2}$. We do not expect any such exact formula for the number of isomorphism types of trees on $n$ vertices. But what are the sharpest asymptotics, or best upper and lower bounds known, as $n \to \infty$?

Draw a tree with 7 vertices. Determine for what \(n\) the tree is \(n\)-connected. Checkpoint \(\PageIndex{4}\) Explain why for any tree removal of a leaf produces another tree. Checkpoint \(\PageIndex{4}\) Find a spanning tree for every graph in Figure 5.2.43.