Primes And A Double Product

First let us fix $a\in\{1,2,\dots,p_n-1\}$ . We observe that $\sqrt[p_n]{p_n}e^{2\pi i b/p_n}$ for $b=1,2,\dots,p_n$ are all the roots of the polynomial $P(x):=x^{p_n}-p_n$ . Moreover, since $a$ is relatively prime to $p_n$ (for $p_n$ is prime and $a<p_n$ ), $\sqrt[p_n]{p_n}e^{2\pi i a b/p_n}$ are also the roots of the same polynomial (not necessarily in the same order). Therefore,

$-1-p_n=P(-1)=\prod_{b=1}^{p_n}\left((-1)^{p_n}-\sqrt[p_n]{p_n}e^{2\pi i b/p_n}\right) =\prod_{b=1}^{p_n}\left(-1-\sqrt[p_n]{p_n}e^{2\pi i b/p_n}\right),$ (notice that $p_n$ is odd, so $(-1)^{p_n}=-1$ .)

Thus, we deduce that

$\prod_{a=1}^{p_n-1}\prod_{b=1}^{p_n} \left(1+\sqrt[p_n]{p_n}e^{2\pi i b/p_n}\right) =\prod_{a=1}^{p_n-1}(1+p_n) =(1+p_n)^{p_n-1}.$

Now, when $a=p_n$ ,

$\prod_{b=1}^{p_n}\left(1+\sqrt[p_n]{p_n}e^{2 i p_n b/p_n}\right) =\prod_{b=1}^{p_n}(1+\sqrt[p_n]{p_n}) =(1+\sqrt[p_n]{p_n})^{p_n}.$

Hence, $f(n)=\prod_{a=1}^{p_n}\prod_{b=1}^{p_n}(1+\sqrt[p_n]{p_n}e^{2\pi i ab/p_n}) =(1+p_n)^{p_n-1}(1+\sqrt[p_n]{p_n})^{p_n}.$

Now it is simple to compute

$\begin{aligned} \sum_{n=1}^{2016}\left( \frac{\sqrt[p_n]{(1+p_n)f(n)}}{1+\sqrt[p_n]{p_n}} -p_n +n\right) &= \sum_{n=1}^{2016}\left( \frac{\sqrt[p_n]{(1+p_n) (1+p_n)^{p_n-1}(1+\sqrt[p_n]{p_n})^{p_n}}}{1+\sqrt[p_n]{p_n}} -p_n +n\right)\\ &= \sum_{n=1}^{2016}\left( \frac{ (1+p_n)(1+\sqrt[p_n]{p_n})}{1+\sqrt[p_n]{p_n}} -p_n +n\right)\\ &= \sum_{n=1}^{2016}\left( 1+p_n -p_n +n\right)\\ &=\sum_{n=1}^{2016}(1+n) =2016+\frac{2016\cdot 2017}2 =2035152. \end{aligned}$