chromatic number proof for $K_n$ So for proving chromatic number of $K_n$ is $n$, and I use the fact that the chromatic polynomial for $K_n$ is $\frac{k!}{(k-n)!}$, is it correct to state $n$ is the minimum number such that the chromatic polynomial has a value greater than $0$?
Negative factorials should not be a problem if I state that right?
 A: Well think about what the chromatic polynomial is. Here you are GIVEN k colors and your chromatic polynomial gives you the number of ways of properly coloring your graph using these colors. 
The chromatic number however is the MINIMUM number of colors needed to color your graph. So sure, if you were given the chromatic polynomial, you COULD find the smallest number n to make this polynomial defined, but I don't think it's necessary. 
Just think about the complete graph. Since every pair of vertices is connected by an edge, then every vertex needs to be a different color, so the chromatic number is the number of vertices: n. No need to bring in the chromatic polynomial
A: The chromatic polynomial for $K_n$ is $P(K_n; t) = t^{\underline{n}} = t (t - 1) \ldots (t - n + 1)$ (a falling factorial power), then the minimal $t$ such that $P(K_n; t) \ne 0$ is $n$. Note that this is a polynomial in $t$ for all $n \ge 1$.
The falling factorial power can be expressed as $t^{\underline{n}} = \frac{t!}{(t - n)!}$, and it is usual to consider factorials of negative numbers to be infinite (this is consistent with the gamma function, $\Gamma(z) = \int_0^\infty u^{z - 1} e^{-u} du$, it is simple to check that for integer $n > 0$ it is $\Gamma(n) = (n - 1)!$). But that takes too far.
