Sum of digits

By the divisibility rules , we know that the sum of the digits divided by 9 will leave the same remainder as when the number itself is divided by 9.

Note that $2016$ is a multiple of 9, so the sum of the digits of $P,$ $Q,$ must be a multiple of 9. By the same logic, since $Q$ is a multiple of 9, so is $R$ and then so is $S.$

We know that $S$ is a positive integer (iteratively summing the digits can never result in 0), but it could be $9, 18, 27, 36, \ldots$

However, note that the sum of the digits of a number is less than 10 times the number of digits it has. Thus, $Q < 10 \cdot \lceil \log_{10}2016^{2016}\rceil = 10 \cdot \lceil 2016 \log_{10}2016\rceil < 10^5.$

Since $Q < 10^5,$ it has at most 5 digits. Thus, $R < 10 \cdot 5 = 50,$ so the maximum sum of the digits of $R$ is $4+9 = 13,$ so $S \le 13.$ But since we determined that $S$ is a positive multiple of 9, $S = \boxed{9}.$

Remark: There are obviously tighter bounds that can be achieved throughout this solution, but they're not needed to solve this problem.