100 solutions, 700 upvotes problem

We have The Binomial Theorem as $(a+b)^n=\displaystyle \sum_{n=0}^{2016} \binom{n}{k} a^{n-k}b^k,$ And similarly when $a=1$ : $(1+b)^n=\displaystyle \sum_{n=0}^{2016} \binom{n}{k} b^k.$ We use it to investigate the required expression as follows: $\begin{aligned} \dfrac{\binom{2016}{0}+2\binom{2016}{1}+\ldots+2^{2016}\binom{2016}{2016}}{\binom{2016}{0}+\binom{2016}{1}+\ldots\binom{2016}{2016}}=&\dfrac{\displaystyle \sum_{n=0}^{2016} 2^k \binom{2016}{k}}{ \displaystyle \sum_{n=0}^{2016} \binom{2016}{k}}=& \dfrac{(1+2)^{2016}}{(1+1)^{2016}} \\ =& \dfrac{3^{2016}}{2^{2016}} \\ =& \left(\dfrac{3}{2}\right)^{2016} \end{aligned}$ Therefore the answer is $\left[\left(\dfrac{3}{2}\right)^{2016}\right]^{\frac{1}{2016}}=\boxed{\dfrac{3}{2}}$ .