Who's up to the challenge? 32

First we observe $\displaystyle\int_{a}^{b} \{nx\}dx = \int_{a}^{b} \{x\}dx$ , so that the 2016 can be disregarded.

Then, on $(0,1), \{x\} = x$ , further reducing the integral to just $\displaystyle\int_{0}^{1} x(x-x^2)^5dx = \int_{0}^{1}x^6(1-x)^5dx$

By integrating by parts many times, we observe a pattern:

$\displaystyle\int_{}^{} x^6(1-x)^5dx = -\frac{x^6(1-x)^6}{6} - \frac{x^5(1-x)^7 \times 6}{6 \times 7} -$

$\displaystyle\frac{x^4(1-x)^8 \times 6 \times 5}{6\times 7 \times 8} - ... -\frac{(1-x)^{12} \times 6!}{\frac{12!}{5!}}$

Following the Second Fundamental Theorem, all the terms cancel out for $x =1$ and only the last term, say $f(x)$ , at $-f(0)$ matters.

$\displaystyle -f(0) = \boxed{\frac{1}{5544}}$