We're give that the polynomial is $k(k-1) \cdots (k-n+1)$ where $k$ represents how many colors we can use and $n$ is the number of vertices.
For the base case I just used a graph with 1 vertex and said that the chromatic polynomial would just be 1 since I can use any of my colors to color that vertex in. Putting one more vertex there and connecting both together, we have $k$ choices for the first vertex color and $k-1$ for the choice of the second color.
My problem is figuring out the inductive step. We are multiplying until $k-n+1$, where that is supposed to represent us subtracting one from each element of the product each time to indicate that we are adding a new vertex so we have one fewer choice for any vertex we want to add. My confusion mainly stems from
1) What the $k-n+1$ is supposed to represent. Is this just the final terms of subtracting one in this way over and over?
2) How we'd prove this via an inductive method. My thought is to have a graph with polynomial $k(k-1) \cdots (k-n+1)$ and then add another vertex that connects to all other vertices. This would mean that we would add another term for the product, but I'm not too sure how to simplify things from there.
Yes, that's right. Whenever we connect the new vertex to every other vertex (say there are $m$ of them), we have $k-m+1$ options for the new vertex.
First of all, in base case, chromatic polynomial is $k$, not $1$ since we can use any of $k$ colors to color one vertex.
Now, suppose inductively that for clique $K_n$ with $n$ vertices, we have chromatic polynomial $k(k-1)...(k-n+1)$. Then for a clique with $n+1$ vertices, we just add one vertex to $K_n$ and connect it to every other vertex. Then this new vertex have $(k-n)$ options because $n$ colors are used for vertices of $K_n$ already. Now, by induction hypothesis, we know that $K_n$ has $k(k-1)...(k-n+1)$ different colorings. Therefore, $K_{n+1}$ has chromatic polynomial $k(k-1)...(k-n+1)(k-n)$ so by induction, argument holds for all $n$.