Ramanujan's nested radical is great! (3)

Let: $a$ $=$ $\sqrt [k]{2016\sqrt [k]{2016\sqrt [k]{2016\sqrt [k]{...}}}}$

$a^k$ $=$ $2016$ $\sqrt [k]{2016\sqrt [k]{2016\sqrt [k]{...}}}$

$a^k$ $=$ $2016$ $a$

$a^{k-1}$ $=$ $2016$

$a$ $=$ $\sqrt [k-1]{2016}$

$\prod_{k=2}^{2017}$ $\sqrt [k-1]{2016}$ $=$ $\sqrt [1]{2016}$ $\sqrt [2]{2016}$ $\sqrt [3]{2016}$ *... $\sqrt [2016]{2016}$

$=$ $2016^{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}$

$=$ $2016^{\sum_{n=1}^{2016} \frac{1}{n}}$

$Thus, x=2016$

$1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dots=\dfrac{1}{1-\frac{1}{n}}$

$1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dots=1+\dfrac{1}{n-1}$

$\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}+\dots=\dfrac{1}{n-1}$

$\Rightarrow \large x^{\frac{1}{n-1}}=x^{\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\dots}$

$\Rightarrow \large x^{\frac{1}{n-1}}=x^{\frac{1}{n}(1+\frac{1}{n}(1+\frac{1}{n}(1+\dots))}$

$\Rightarrow \large x^{\frac{1}{n-1}}=\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\sqrt[n]{...}}}}$

$\Rightarrow \displaystyle \prod_{k=2}^{2017}\sqrt[k]{2016\sqrt[k]{2016\sqrt[k]{2016\sqrt[k]{2016\sqrt[k]{...}}}}}=\displaystyle \prod_{k=2}^{2017}2016^{\frac{1}{k-1}}$

$\displaystyle \prod_{k=2}^{2017}2016^{\frac{1}{k-1}}$

$= 2016^{1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2016}}$

$=2016^{\left(\displaystyle \sum_{n=1}^{2016}\frac{1}{n}\right)}$

$\Rightarrow x=2016$