Down With "Plug and Chug"

As suggested, take $f(f(x))$ .

$f(f(x)) \rightarrow f\left(\frac{x + 1}{x-1}\right)$

Substituting $\frac{x-1}{x-1} =1$ , we get: $f\left(\frac{x + 1}{x-1}\right) = \frac {\frac{x + 1 + (x - 1)}{x-1}}{\frac{x + 1 - (x - 1)}{x-1}}$

We can then cancel out the common denominator and simplify the fraction.

$f(f(x)) = \frac{x + 1 + (x - 1)}{x + 1 - (x - 1)} = \frac{2x}{2} = x$

Now, we have determined that $f(f(x)) = x$ , and plugging it into the expression $f(f(f(f(2016))))$ yields $\boxed{2016}$ .

$f(x)=\frac{x+1}{x-1}=\frac{(x-1)+(2)}{x-1}=1+\frac{2}{x-1}$

$f(f(x))=1+\frac{2}{(1+\frac{2}{x-1})-1}=1+\frac{2}{\frac{2}{x-1}}=1+x-1=x$

$x=f(f(x))=f(f(f(f(x))))$

$f(f(f(f(2016))))=\boxed{2016}$

As it is componendo - dividendo we can take x/1, then substitute x equal to 2016.