Let $n\gt k\gt1$ with $n$ even and k odd. Make a $k$-regular graph $G$ by putting $n$ vertices in a circle and connecting each vertex to the exact opposite vertex and $\frac{k-1}{2}$ closest vertices on either side. Show that for all $u,v$ there are $k$ internally disjoint $u,v$-paths.
I know there is a theorem
A graph is $k$-connected if and only if it contains $k$ internally vertex-disjoint paths between any two vertices.
But is there any direct way to prove it has exactly $k$ internally disjoint $u,v$-paths.