Square Root Calculations

$\begin{aligned} a & = (4+\sqrt{15})(\sqrt 5-\sqrt 3)\color{#3D99F6}\sqrt{4-\sqrt{15}} & \small \color{#3D99F6} \text{Note that }(\sqrt 5-\sqrt 3)^2 = 8 - 2\sqrt{15} \\ & = (4+\sqrt{15})(\sqrt 5-\sqrt 3)\color{#3D99F6}\sqrt{\frac {(\sqrt 5-\sqrt 3)^2}2} \\ & = (4+\sqrt{15})\times \frac {(\sqrt 5-\sqrt 3)^2}{\sqrt 2} \\ & = (4+\sqrt{15})\times \frac {8-2\sqrt{15}}{\sqrt 2} \\ & = (4+\sqrt{15})(4-\sqrt{15})\sqrt 2 \\ & = (4^2-15)\sqrt 2 \\ & = \sqrt 2 \end{aligned}$

Therefore, $f(a) = f(\sqrt 2) = \left((\sqrt 2)^4 + \sqrt 2 \cdot \sqrt 2 - 7\right)^{2016} = (4+2-7)^{2016} = (-1)^{2016} = \boxed 1$ .