Every tournament has a Hamiltonian path (Rédei's theorem)

Step 1 (extremal choice). Among all directed paths in the tournament, choose one, $P: v_1 \to v_2 \to \cdots \to v_k$, with the largest possible number of vertices $k$; such a maximum exists because there are only finitely many vertices. Suppose, for contradiction, that $k < n$, and let $u$ be a vertex not on $P$.

Step 2 (case split at the two ends). Since the tournament has exactly one of $u \to v_1$ or $v_1 \to u$: if $u \to v_1$ held, prepending $u$ would give the longer path $u \to v_1 \to \cdots \to v_k$, contradicting the maximality of $k$; hence $v_1 \to u$ holds. Symmetrically, exactly one of $v_k \to u$ or $u \to v_k$ holds: if $v_k \to u$ held, appending $u$ would give a longer path, contradicting maximality; hence $u \to v_k$ holds.

Step 3 (locate an insertion point). We now know $v_1 \to u$ and $u \to v_k$. Let $j$ be the largest index in $\{1, \dots, k-1\}$ such that $v_j \to u$ holds; this set of indices is nonempty since $j = 1$ works, so by the extremal principle $j$ exists. By maximality of $j$, the arc $v_{j+1} \to u$ does not hold, so the tournament property forces $u \to v_{j+1}$.

Step 4 (contradiction by insertion). Combining $v_j \to u$ with $u \to v_{j+1}$, we insert $u$ between $v_j$ and $v_{j+1}$ to form the path $v_1 \to \cdots \to v_j \to u \to v_{j+1} \to \cdots \to v_k$, which has $k + 1$ vertices, contradicting the maximality of $k$.

Step 5 (conclusion). No such vertex $u$ can exist, so $k = n$ and $P$ is a Hamiltonian path.