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  "body": "\n## Proof\n\nWe take a longest path and show that its last vertex is a leaf.\n\n1. The [[tree|0MJ0JS72DZ]] $T$ is [[connected|0FFAYNMAJZ]] and [[acyclic|0XNN9Z8E5A]]. Let $n \\ge 2$ be its number of vertices.\n2. Choose distinct vertices $u \\ne v$. By connectedness there is a [[walk|0VJZM0F9YM]] $(w_0, \\dots, w_k)$ from $u$ to $v$. Since $u \\ne v$, $k \\ge 1$.\n3. So $w_0 \\sim w_1$, and $(w_0, w_1)$ is a [[path|0JDE56HWQ8]] of length $1$.\n4. A path has pairwise distinct vertices, so its length is at most $n - 1$. Hence there is a path $P = (v_0, v_1, \\dots, v_k)$ of largest length, and $k \\ge 1$ by step 3.\n5. The vertex $v_{k-1}$ is a neighbour of $v_k$, so $\\deg(v_k) \\ge 1$. Let $x$ be any neighbour of $v_k$. We show that $x = v_{k-1}$.\n6. If $x$ is not a vertex of $P$, then $(v_0, \\dots, v_k, x)$ is a path of length $k + 1$. This contradicts the choice of $P$.\n7. So $x = v_i$ for some $i$. Since $x \\sim v_k$ and no vertex is adjacent to itself, $i \\ne k$.\n8. Suppose $i \\le k - 2$. Consider the closed walk $C = (v_i, v_{i+1}, \\dots, v_k, v_i)$ of length $m = k - i + 1 \\ge 3$.\n9. The vertices of $C$ after the first one are $v_{i+1}, \\dots, v_k, v_i$. They are pairwise distinct, since $P$ is a path.\n10. The edges $v_j v_{j+1}$ for $i \\le j \u003c k$ are distinct, because their vertex pairs differ. The last edge $v_k v_i$ differs from each of them, because $\\set{v_k, v_i} = \\set{v_j, v_{j+1}}$ would force $i = k - 1$.\n11. So $C$ is a [[circuit|0PD57PSHNR]] whose vertices after the first are pairwise distinct. That is, $C$ is a [[cycle|07CT9RDY7Q]] of $T$, which contradicts acyclicity.\n12. Hence $i = k - 1$, and $x = v_{k-1}$. So $v_k$ has exactly one neighbour.\n13. By the definition of the [[degree|0T0F6CCS9W]], $\\deg(v_k) = 1$. So $v_k$ is a vertex of degree $1$ of $T$.\n\n## Idea\n\nWalk along a longest path to its end. The last vertex cannot have a neighbour off the path, or the path would grow. It cannot have a second neighbour on the path either, or the tree would contain a cycle.\n",
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