I'm very confused as to why the degree of any chromatic polynomial of a simple graph $G$ that has $n$ vertices is $n$. I've tried using strong induction on the number of edges, but I end up getting something absurd.
Let $n$ be the number of vertices and $m$ be the number of edges. Consider a simple graph with n vertices and no edges. Then clearly the chromatic polynomial $P_G(x) = x^n$.
Now suppose that $G$ has at least one edge and let $e$ be an edge of $G$. Then by the Deletion-Contraction Theorem, $P_G(x) = P_{G-e}(x) - P_{Ge}(x)$. But both graphs $G-e$ and $Ge$ has fewer edges than $G$ itself, so by the inductive hypothesis, these two graphs with fewer edges have a chromatic polynomial of degree $n$.
The part that confuses me the most is that in general, $$\deg(f(x) - g(x)) ≤ \max\{\deg(f(x)), \deg(g(x))\}.$$ In this case, the degree of $P_G(x)$ is less than or equal to $n$. So what if you get chromatic polynomials for those two graphs with fewer edges both whose coefficients of $x^n$ is just simply equal to one? Then $x^n$ cancels out in the chromatic polynomial of the graph $G$ itself, which contradicts the fact that the degree of the chromatic polynomial of any simple graph with n vertices is $n$.
Can anybody point out to me what I'm doing wrong? Thank you in advance.