 Solution

Induct using the same deletion-contraction recurrence for the chromatic polynomial. The edgeless graph has polynomial $x^n$, so its leading coefficient is one and the others are zero. All nonempty-vertex chromatic polynomials have zero constant term, since no coloring uses zero colors.

For an existing edge, write the alternating coefficient lists of the deletion and simplified contraction as $a_i^{\prime }$ and $a_i^{\prime \prime }$. The contraction has $n-1$ vertices, so subtracting its polynomial changes its leading sign by one place. Thuswith nonexistent indices understood as zero. Both summands are nonnegative by induction. The leading coefficient is $a_0=1$, and the next isThis proves the alternating expansion and nonnegativity of every $a_i$. The coefficient assertion for $a_1$ is relevant when $n\ge 2$; the one-vertex graph simply has polynomial $x$.