Prove that in a tree with maximum degree $k$, there are at least $k$ leaves

Let me give you a hint.

Start off by proving that every tree with at least two vertices must have at least 2 leaves.

Now, if the maximum degree is $k$, then there is a vertex $v$ of degree $k$. Consider the graph obtained by deleting $v$ from your tree. What does it look like? Prove that it is a collection of $k$ non-empty trees.

Consider each of these components. If the component has only one vertex, then this vertex was a leaf in the original graph: its only neighbor was $v$.

If the component has two or more vertices, then by the first claim it must have two leaves. Of these, only one could have been connected to $v$, since otherwise they form a cycle; hence at least one of them must have been a leaf in the original tree.

So, we get at least one leaf from each of the $k$ components of the new graph.

Let the maximum degree be $k$. Start at a vertex of degree $k$ and travel away from it. There are $k$ possible choices of starting edge, each of which leads to at least one leaf. The leaves must be distinct, for otherwise there would be a cycle.