Lower bound for The chromatic index of the complete graph of order n, where n is odd. How can I prove that the chromatic index i of the complete graph of order n, where n is odd, is i > n-1.
I found here a construction with n colors, but am having a hard time to prove lower bound.

Hint Assume by contradiction that you can color $K_n$ with $n-1$ colors. Then, at each vertex each color appears exactly once.
Now, pick one color, and draw the subgraph of all edges of this fixed color. This is a subgraph on $n$ vertices, where each vertex has degree 1. But this is not possible by handshaking lemma, since $n$ is odd.
P.S. Alternately, pick one color and show that it only appears at an even number of vertices (i.e. the number of vertices having an incident edge of that color is even). Therefore, this color is missing at one vertex, and that vertex has n-1 edges. [think in terms of sports: n teams tournament, each team plays at most a game per day, colors=days. On day 1 a team must have a bye, and still n-1 games left]
