For my answer so far, I've got something along the lines of:
"K$_n$ is a complete graph if each vertex is connected to every other vertex by one edge. Therefore if n is even, it has n-1 edges (an odd number) connecting it to other edges. Therefore it can't be Eulerian..." which comes from this answer on Yahoo.com.
I guess I want to check that the rest of what's contained in that answer is correct and good to base my answer off of. Any confirmation or help would be appreciated!
From Euler's theorem, graph $G$ is Eulerian iff every vertex has even degree. Hence, for $K_n$ and n´odd, deg(v) are even, so it is Eulerian. For odd n, by Euler's theorem implies that it is not Eulerian.