Prove that a tree with a vertex $v$ of degree $k > 1$ has at least $k$ leaves My assignment is the following:
Prove that if a tree that has a vertex $v$ of degree $k > 1$ has at least $k$ leaves.
$\delta_T(v)$ means the degree of $v$ in the tree $T$
This is the approach I have:
Let there be a tree $T$. By statement we know that there is a vertex of degree $ k> 1 $. This means that there is a vertex that has more than one edge and also has an equal or greater number of edges than any other $ T $ vertex. Let the previous vertex be the vertex $ v '$. Since it has a degree $ k > 1 $, then it is not a leaf.
Because of the Handshaking Lemma, the sum of the degrees of the vertices is twice the number of edges, then:
$$\sum_{v\in V}{\delta_T(v)}=2|E|\ (1)$$ 
By property of trees, we know that $ | E | = | V | -1 $ $ (2) $. 
Replacing $ (2) $ in $ (1) $, we can say that
$$\sum_{v\in V}{\delta_T(v)}=2|V|-2\ (*)$$
If we define $ L $ as the set of all the leaves of $ T $. The degree of all vertices that are not leaves is at least 2. Then:
$$\sum_{v\in V}{\delta_T(v)}=\delta_T(v')+\sum_{v\in L}{\delta_T(v)}+\sum_{v\in V\backslash(L\cup [v'])}{\delta_T(v)}$$
In the case that all the leaves have degree 1 in $ T $ and assuming that the degree of $ v '$ is $ k $:
$$\sum_{v\in V}{\delta_T(v)}\geq k+|L|+2(|V|-|L|-1)$$
$$\sum_{v\in V}{\delta_T(v)}\geq k-|L|+2|V|-2$$
Replacing $ (*) $ in the previous inequality:
$$2|V|-2\geq k-|L|+2|V|-2$$
$$|L|\geq k$$
This translates into the fact that the amount of leaves that exist in the tree must be at least the amount equal to the largest degree that has a vertex in the whole tree.
Is there anything to add to this solution? Maybe something to fix?
Thank you in advance.
 A: Your introduction of $v'$ is a little confusing. I would instead say "Let $v'$ be a vertex of maximal degree in $T$. Since we know there is a vertex of degree $k>1$ in $T$, it follows that $\delta_T(v') \geq k$." Or alternatively, use some notation in addition to words, such as "There is a vertex $v'$ such that $\delta_T(v') \geq \delta_T(v'')$ for any vertex $v''$ of $T$. Since $\delta_T(v)=k$, it follows that $\delta_T(v'') \geq k$."
Note that you can't assume $\delta_T(v')=k$ only that $\delta_T(v') \geq k$, but that does not affect the inequality later in your proof. 
As an aside, I don't think the restriction $k>1$ is necessary. If $k=1$ then the only possibility is the path graph with two vertices, which has at least $1$ leaf.
A: It is a nice complete proof, but what about the sentence " This means that there is a vertex that has more than one edge and also has an equal or greater number of edges than any other T vertex."? The vertex $v'$ need not be the vertex with the maximum degree, and the proof will still be right.
Also the notation $\delta(v)$ is usually used to denote the minimum degree in a graph, so I think it is better to use $d(v)$ 
