I'm attempting exercise 15 from the end of the first section in Diestel's Graph Theory. The exercise asks:
Show that any tree $T$ has at least $\Delta(T)$ leaves. $\Delta(T)$ denotes the maximum degree of any vertex in T. $d(v)$ denotes the degree of vertex $v$
I came up with the following:
$Proof.$ Let $T = (V, E)$ be a tree with an arbitrarily selected root, $r$. Let $L \subseteq V$ denote the set of vertices of degree 1 in T, i.e. the leaf vertices of $T$.
Now let $v \in V - L$ be an inner-vertex of $T$ (i.e. not a leaf), and let $d(v) = \Delta(T)$. As an inner-vertex of the tree, clearly $v$ must lie on some path $rTu$ for some $u \in L$, and as a result, $1$ of the $\Delta(T)$ incident edges of $v$ must lead to $u$. By the definition of a tree, each path from $r$ to a leaf $w \in L$ is uniquely defined. It follows that all $\Delta(T) - 1$ other edges incident to $v$ must belong to unique paths leading to other leaf nodes $w \neq u$, giving us at least $\Delta(T)$ leaves. $\Box$
I'm not convinced. Any suggestions?