Ramanujan's nested radical is great! (4)

Rewrite the above expression :

$\large \dfrac{(\sqrt[0!]{2016^{\pi^{0}}})(\sqrt[1!]{2016^{\pi^{1}}})(\sqrt[4!]{2016^{\pi^{4}}})(\sqrt[5!]{2016^{\pi^{5}}})(\sqrt[8!]{2016^{\pi^{8}}})(\sqrt[9!]{2016^{\pi^{9}}})(\dots)}{(\sqrt[2!]{2016^{\pi^{2}}})(\sqrt[3!]{2016^{\pi^{3}}})(\sqrt[6!]{2016^{\pi^{6}}})(\sqrt[7!]{2016^{\pi^{7}}})(\sqrt[10!]{2016^{\pi^{10}}})(\sqrt[11!]{2016^{\pi^{11}}})(\dots)}$

$=\large \exp_{2016} (1+\pi-\frac{\pi^{2}}{2!}-\frac{\pi^{3}}{3!}+\frac{\pi^{4}}{4!}+\frac{\pi^{5}}{5!}-\frac{\pi^{6}}{6!}-\frac{\pi^{7}}{7!}+\dots)$

$=\large \exp_{2016} (\sin(\pi)+\cos(\pi))$

$=\large \exp_{2016} (-1)$

$\Rightarrow x=-1$

Notation : $\exp_A (B)$ denotes the exponential function $A^B$ .

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Consider the power series for $\cos(x)$ and $\sin(x)$ :

$\sin(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\dots$

$\sin(\pi)=\pi-\frac{\pi^{3}}{3!}+\frac{\pi^{5}}{5!}-\frac{\pi^{7}}{7!}+\dots$

$\cos(x)=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\dots$

$\cos(\pi)=1-\frac{\pi^{2}}{2!}+\frac{\pi^{4}}{4!}-\frac{\pi^{6}}{6!}+\dots$

$\sin(\pi)+\cos(\pi)=1+\pi-\frac{\pi^{2}}{2!}-\frac{\pi^{3}}{3!}+\frac{\pi^{4}}{4!}+\frac{\pi^{5}}{5!}-\frac{\pi^{6}}{6!}-\frac{\pi^{7}}{7!}+\dots$

$\sin(\pi)+\cos(\pi)=0-1=-1$