I'm reading Barbeau's Polynomials.
I've done the following: Taking an arbitrary chessboard $C$ with some of the squares forbidden (with $n$ being the number of squares and $F$ the number of free squares), I'd have:
$$C=\binom{n-F}{0}t^0+\binom{n-F}{1}t^1+\binom{n-F}{2}t^2+\dots+\binom{n-F}{c}t^c$$
Then I guess that $C_1$ would be:
$$C_1=\binom{n-F-1}{0}t^0+\binom{n-F-1}{1}t^1+\binom{n-F-1}{2}t^2+\dots+\binom{n-F-1}{c}t^c$$
For $C_2$, I guess that I'd have to speak about other number of free squares, $f$ for example:
$$C_2=\binom{n-f}{0}t^0+\binom{n-f}{1}t^1+\binom{n-f}{2}t^2+\dots+\binom{n-f}{c-1}t^{c-1}$$
Now, to show that $C=C_1+C_2$, I guess that I'd need to sum the terms this way:
$$\binom{n-F-1}{x}t^x+ \binom{n-f}{x}t^x=\frac{(n-F-1)!}{x!(n-F-1-x)!}t^x+\frac{(n-f)!}{x!(n-f-x)!}t^x$$
But from here, I'm stuck. I'm not sure how to proceed.
You hardly need any formulas at all, just focus on the $r_k$.
Consider a valid configuration on board $C$. Either there is a rook on $S$ or there is no rook on $S$.
If there is a rook on $S$, there cannot be any rook in the entire row or column containing $S$, so the remaining $k-1$ rooks are determined by the placement of $k-1$ rooks in $C_2$.
On the other hand the valid configurations without a rook on $S$ correspond to the valid placements of $k$ rooks on $C_1$.
So $r_k(C)=r_k(C_1)+r_{k-1}(C_2)$.
Now multiply with $t^k$ and sum over $k\geq 1$ to get the desired relation between the rook polynomials.