Let $T$ be a tournament directed graph (round robin). Prove that there is an odd number of Hamiltonian paths in the graph 
Let $T$ be a tournament directed graph (round robin). Prove that there is an odd number of Hamiltonian paths in the graph.

I am aware that this question may be considered a duplicate of this one: The number of Hamiltonian paths in a tournament is always odd?. However, that question has no answer and the only interaction is a link to a MathOverflow post https://mathoverflow.net/questions/232751/the-number-of-hamiltonian-paths-in-a-tournament which cites Rédei's theorem without proof, and with a reference that is now dead link and not archived under WayBack Machine.
I am also aware of Rédei's theorem, in the form that is saying that there always exists a directed Hamiltonian path in a tournament. https://everysnap.blogspot.com/2010/07/redeis-theorem.html
However, I am unable to solve the problem / the strong theorem, so, I'm searching for an elementary solution.
 A: The MathOverflow post has several alternative links in the comments, and in particular Berge's proof (Theorem 6 in section 10.2 of Graphs and Hypergraphs) seems fairly nice.
First, Berge proves a theorem about complements. (Here, if $G$ is a directed graph, $\overline{G}$ is the directed graph which has all possible directed edges $G$ does not have; if $G$ has no edge in either direction between $u$ and $v$, then $\overline G$ has an edge in both directions.) I will write $h(G)$ for the number of Hamiltonian paths in $G$.

Theorem 1. If $G$ is any $n$-vertex directed graph with $n>1$, then $h(G) \equiv h(\overline G) \pmod 2$.

Let $s(H)$ denote the number of ways to number the vertices of $H$ as $1$ through $n$ such that all edges are of the form $(i,i+1)$. Then $s(H)=0$ unless $H$ is a union of paths; if $H$ has $k$ path components, then $s(H) = k!$. Therefore $s(H) \equiv 1 \pmod 2$ only if $H$ is a directed path. In particular, $$h(G) \equiv \sum_{E(H) \subseteq E(G)} s(H) \pmod 2$$ where the sum is over all spanning subgraphs of $G$.
Meanwhile, if we take a Hamiltonian path in $\overline G$ and number the vertices of $G$ along that path, then no edge of $G$ is of the form $(i,i+1)$, because all such edges are in $\overline G$. So $n! - h(\overline G)$ counts the number of labelings in which at least one edge of $G$ is of the form $(i,i+1)$. By the principle of inclusion-exclusion, we get
$$
   n! - h(\overline G) = \sum_{E(H) \subseteq E(G)} (-1)^{|E(H)|} s(H) \equiv \sum_{E(H) \subseteq E(G)} s(H) \pmod 2
$$
and as we saw earlier, this means $n! - h(\overline G) \equiv h(G) \pmod 2$. For $n>1$, this gives $h(G) \equiv h(\overline G) \pmod 2$.
Now we are ready to prove:

Theorem 6. If $T$ is a tournament, $h(T) \equiv 1 \pmod 2$.

The idea is to check that the parity of $h$ does not change if we reverse an edge. Thus, the parity is the same as the number of Hamiltonian paths in a transitive tournament, which is $1$.
Let $e$ be an edge of $T$; let $f$ be the reverse of $T$. We define four directed graphs from $T$: $G_0 = T - e$, $G_e = T = G_0 + e$, $G_f = G_0 + f$, and $G_{ef} = G_0 + e + f$. So these agree everywhere except about which of edges $e$ and $f$ they have.
Note that $\overline{G_{ef}}$ gives us $G_0$ with all of its edges reversed, and reversing all the edges does not change the number of Hamiltonian paths. So by Theorem 1, $h(G_{ef}) \equiv h(G_0) \pmod 2$.
We have $h(G_{ef}) - h(G_0) = (h(G_e) - h(G_0)) + (h(G_f) - h(G_0))$. Here, the LHS counts Hamiltonian paths that use edge $e$ or $f$, and the RHS has two terms counting each of those cases separately. Therefore
$$
   h(G_e) + h(G_f) = h(G_{ef}) + h(G_0) \equiv 0 \pmod 2
$$
from which we get $h(G_e) \equiv h(G_f) \pmod 2$.
