Let’s place the rooks one by one, indexing them starting with 0. Let \(P(n)\) denote the probabiity that the rook \(n\) drops to a safe square given that the previous \(n-1\) rooks were placed safely.

Obviously \(P(0)\) is 1. For \(P(1)\), we have \(64 - 15\) safe squares out of 63 remaining, as the first placed rook puts to danger 15 squares. The next rook will have to be dropped into one of \(65 - 15 - 13\) safe squares, as the second rook put to danger 13 squares in addtion to 15 controlled by the first, and so on. We apply the chain rule to compute \(P(n)\) in the general:

\[P(n) = \prod_{k= 1}^n \frac{64-\sum_{r=0}^{k-1} 15 - 2\cdot r}{64 - k}\]

Therefore,

\[P(7) = \frac{49}{63} \cdot \frac{36}{62} \cdot \frac{25}{61} \cdot \frac{16}{60} \cdot \frac{9}{59} \cdot \frac{4}{58} \cdot \frac{1}{57}\]

An alternative solution would be to first count the different ways of choosing where to put the rooks ensuring they are safe, and then divide this by the total number of ways to place 8 rooks without caring for keeping them safe. In numbers,

\[\frac{64\cdot 49 \cdot 36 \cdot 25 \cdot 16 \cdot 9 \cdot 4 \cdot 1}{64\cdot 63\cdot 62\cdot 61\cdot 60\cdot 59\cdot 58\cdot 57}\]