TITLE: Proof in graph theory: maximum degree and number of leaves.
QUESTION [8 upvotes]: I have to prove that if $v$ is a vertex of maximum degree in a tree and the degree of $v$ is $k$, then the tree has at least $k$ vertices of degree $1$.
I think I get why that is true: If $v$ has $k$ neighbors, then these may or may not have neighbors. If they are leaves, then we are done. If they do have neighbors, then eventually we will reach the leaves, since the tree is finite. So it is obvious that the tree has at least $k$ vertices of degree $1$. But how do I prove it formally?

REPLY [3 votes]: If $v$ is a vertex of degree $k$ in the tree $T,$ then $T-v$ is a forest of $k$ trees $T_1,\dots,T_k,$ one for each neighbor of $v$ in $T.$ If $T_i$ consists of a single vertex, then that vertex was a leaf in $T.$ If $T_i$ contains more than one vertex, then $T_i$ has at least two leaves; at least one of them was not a neighbor of $v,$ so it was also a leaf in $T.$