Hamiltonian path in directed graph
Proof by induction
Basis. All complete oriented digraphs with two vertices have Hamiltonian path - it's obvious.
Induction step. Let's assume that complete oriented digraph $G_{n-1}$ with $(n - 1)$ vertices contains Hamiltonian path $(s, ..., t)$ from vertex $s$ to vertex $t$. We'll show how to build Hamiltonian path for complete oriented digraph $G_n$, which is constructed from $G_{n-1}$ by adding a vertex $p$.
Case 1. The $G_n$ has an edge $(p, s)$. The Hamiltonian path in $G_n$ will be $(p, s, ..., t)$.
Case 2. The $G_n$ has an edge $(s, p)$. Then we need to scan the Hamiltonian path $(s, ..., t)$, trying to find an edge $(v_{i-1}, v_i)$, such that the graph $G_n$ contains edges $(v_{i-1}, p)$ and $(p, v_i)$. There are two subcases:
Subcase 2.1. We can find such a pair - then the new Hamiltonian path will consist of three parts - $(s, ..., v_{i-1})$, $(v_{i-1}, p, v_i)$ and $(v_i, ..., t)$.
Subcase 2.2. We can't find such a pair - it means, that all the edges from vertices on the path $(s, ..., t)$ to the additional vertex $p$ are oriented to this vertex. The new Hamiltonian path will be $(s, ..., t, p)$.