The power of Hölder Continuity

Replace $x \to x + h$ and $y \to x$

$\Rightarrow |f(x+h) - f(x)| \leq (x + h - x)^{2016} \Rightarrow |f(x+h) -f(x)| \leq h^{2016}$

$\Rightarrow -h^{2016} \leq f(x+h) - f(x) \leq h^{2016}$

$\Rightarrow -h^{2015} \leq \dfrac{f(x+h) - f(x)}{h} \leq h^{2015}$

$\Rightarrow \displaystyle \lim_{h\to 0} -h^{2015} \leq \lim_{h \to 0} \dfrac{f(x+h) -f(x)}{h} \leq \lim_{h\to 0} h^{2015}$

$\Rightarrow 0 \leq f'(x) \leq 0 \Rightarrow f'(x) = 0$

Hence $f(x)$ is a constant function.

$\Rightarrow f(1) = f(2) = \boxed{2016}$

It is easy to see that $|f(2)-f(1)|\leq \sum_{i=1}^n|f(1+\frac{i}{n})-f(1+\frac{i-1}{n})|\leq \sum_{i=1}^n (\frac{1}{n})^{2016}=\frac{1}{n^{2015}},$ for any natural number $n.$ Then make $n\rightarrow \infty.$ So, we get that $|f(2)-f(1)|\leq 0.$ Therefore, $f(2)=f(1)=\boxed{2016}.$