A vertex $v$ in digraph $D$ is said to be reachable. from a vertex $u$ in $D$ if $D$ contains a $u-v$ path.
Let $D$ be a digraph and for each vertex $u$ of $D$, let $R(u)$ be the set of vertices reachable from $u$ and let $r(u)=|R(u)|$. Since $u∈R(u)$ for every vertex $u∈D$, it follows that $r(u)≥1$. Prove that if $r(x)≠r(y)$ for every 2 distinct vertices $x$ and $y$ of $D$ then $D$ contain a Hamiltonian path.
Let $D$ be a digraph of order $n$ and label vertices of $D$ $\{v_1, v_2, ... , v_m \}$ such that $r(v_i)=i$ for $1\leq i\leq n$. So there are $i$ vertices that is reachable from $v_i$. So the vertex $v_n$ is reachable to it self and $n-1$ other vertices. So there is a path from $v_n$ to $v_{n-1}$.
Claim: $v_{n-1}$ also reachable to it self and $n-2$ other vertice. Proof: assume that there is a vertex $w$ such that there is no path from $v_{n-1}$ to $w$.
Case 1: if $w=v_n$ then we have a path between $v_{n-1}$ and $w$
Case 2: if $w \not =v_n$ then there is a path from $v_n$ to $w$, thus there must be a path between $v_{n-1}$ and $w$
both cases lead to $v_{n-1}$ also reachable to it self and $n-2$ other vertice
Claim: the path between $v_n$ and $v_{n-1}$ must be an arc. Proof: Assume that there is some vertex $v_k$ lie on the $v_n -v_{n-1}$ path, then $v_k$ is reachable to $v_n, v_{n-1}$ and all other vertices, which cause $v_k$ to have bigger index then $v_n$ , but $v--N$ have the highest index, thus this is a contradiction and the path between $v_n$ and $v_{n-1}$ must be an arc.
Repeating the argument that the path between $v_{n-1}$ and $v_{n-2}$ must be an arc and so on, we have a directed halmiltonian path $v_n \to v_{n-1} \to \dots \to v_1$
Does my argument sound acceptable?