Discrete Mathematics - Trees Question: Prove that if a tree has a node of degree $n$, it has at least $n$ nodes of degree $1$.


My answer: From each of $n$ edges adjacent to the node of degree $n$ a path starts. Each of these paths eventually ends in a leaf. Since the tree has no cycles, then all these leaves are different, hence the tree has at least $n$ leaves, and degree of a leaf is $1$, thus, the tree has at least $n$ nodes of degree $1$.

I'm not sure about the strength of this argument. Can anybody give me some idea about my answer please.
Let v be a node with degree n in a finite graph.
Let v(k) be the k-th vertex for which v,v(k) is an edge.
Let p(k) be a path of maximal length from v through v(k).
As the path has no loops and is finite it will end in a leaf.  

Now prove there are at least n leaves.
Suppose the tree has $k$ vertices.Then it has $k-1$ edges.

Assume on the contrary that there are at most $n-1$ leaves. Then the number of edges is at least:
$$\frac{n+(n-1)+2(k-n)}{2}=\frac{2k-1}{2} &gt; k-1$$
Contradiction.