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I want to prove the statement

Every tree has at least △(G) leaves

where △(G) is the highest vertex degree. I've seen a proof for this statement, but it seemed overly complicated, so I wanted to construct a simpler proof.

I thought of a proof that goes as follows: If we let $v$ be this vertex of highest degree. So v has △(G) neighbours, and as G is a tree, if we follow a path starting at v and going through each of these neighbours, we eventually find a leaf, hence we have △(G) leaves.

Is this a valid proof? I'm not fully sure how to formulate my idea, but it basically entails starting at $v$ and then going to a neighbour and following a path until we find a leaf, which we know will exist in the path eventually as G is a tree, and we have at least △(G) such paths.

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"we eventually find a leaf" is true, but needs a small argument, involving the fact that we have no cycles. Maybe this is already a proposition in the text you're using, so you can refer to that in that case.

Argue also that the different paths are vertex-disjoint (except for the starting vertex $v$) because otherwise again we'd have a cycle. It's best to make this explicit. In particular then the leaf-vertices are distinct.

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