That prove is easy for n odd.

All k edges of same color of a hypothetical solution connect at most 2k different nodes.

But n is odd, hence one node must skip that color.

Or

A complete graph Kn has n*(n-1)/2 edges and (n-1)/2 is the max number of edges of same color for n odd.

So n is the min of needed colors.

And I've already shown an easy construction with n colors for all Kn (odd and even).