Because $f(x)$ is a $1^{st}$ degree polynomial, $f(f(f(\cdots f(x))))$ is a $1^{st}$ degree polynomial too. This means that the given equation has only one root. Clearly, $f(x)=x$ satisfies the given equation. Therefore, the solution to $f(x)=x$ also satisfies the given equation. Hence, the desired result is $$f(x)=x\implies 2016x-2015=x \implies x=\boxed{1}$$