 Solution

Write $V=V(G)$ and $E=E(G)$. For $F\subseteq E$, let $c(F)$ denote the number of connected components of a graph of the spanning subgraph $(V,F)$, including isolated vertices. The inclusion-exclusion principle, applied to the unwanted events that the two endpoints of an edge receive the same colour, gives the chromatic polynomial inclusion-exclusion formulaTo justify the exponent, imposing equal colours across all edges in $F$ forces a single colour on each connected component of a graph of $(V,F)$, with independent choices between its connected components of a graph. Thus the corresponding count at $q$ colours is $q^{c(F)}$. Equality for all positive integers proves the polynomial identity.

Fix an index $i$ in the stated range. A subset $F$ contributes to the coefficient of $x^{n-i}$ exactly when $n-c(F)=i$. If $(V,F)$ contains a graph cycle of length $s$, the hypothesis gives $s\ge \ell$. All its $s$ vertices lie in one connected component of a graph. For any connected component of a graph $D$, its contribution to $n-c(F)$ is $|V(D)|-1$, soThis contradicts $i\le \ell -2$. Therefore every contributing $(V,F)$ is a forest. The forest identity $|F|=n-c(F)$ then gives $|F|=i$.

Conversely, every $i$-edge subset is a forest, since a graph cycle in it would have at most $i<\ell$ edges. It therefore has $c(F)=n-i$ and contributes exactly $(-1{)}^i$. There are $\left(\genfrac{}{}{0ex}{}{m}{i}\right)$ such subsets, by the definition of a binomial coefficient. ConsequentlyHere the coefficient notation is understood within the defined indices $0\le i\le n$; if the displayed range is extended past $n$, the coefficients beyond the constant term can be set to zero. For a forest, $m\le n-1$, so the binomial coefficients there are also zero. The key reason girth determines initial chromatic coefficients is that a graph cycle can affect only exponents whose loss of connected components of a graph is at least $\ell -1$.