Logarithm

$\begin{aligned} \sum_{i=3}^{2015} \frac{1}{\log_{i^k}2015!} & > 2015 \\ \Rightarrow \sum_{i=3}^{2015} \frac{k \log{i}}{\log 2015!} & > 2015 \\ \frac{k}{\log 2015!} \sum_{i=3}^{2015} \log{i} & > 2015 \\ \frac{k}{\log 2015!} \log \left( \prod_{i=3}^{2015} i \right) & > 2015 \\ \frac{k}{\log 2015!} \log \left(\frac{2015!}{2} \right) & > 2015 \\ \frac{k}{\log 2015!} \left(\log{2015!}-\log{2} \right) & > 2015 \\ k \left( 1 - \frac{\log{2}}{\log{2015!}} \right) & > 2015 \\ k & > \frac{2015}{1 - \color{#3D99F6}{\frac{\log{2}}{\log{2015!}}}} \quad \quad \small \color{#3D99F6} {\text{Since } \frac{\log{2}}{\log{2015!}} \approx 0} \\ \Rightarrow k > \color{#3D99F6}{2015} \end{aligned}$

Therefore, the minimum integer $k = \boxed{2016}$ .

Equivalent to $\sum_{x= 3}^{2015}{\log_{2015!}{x^k}} = \log_{2015!}{(2015!)^k} - \log_{2015!}{2^k}$

$= k - \log_{2015!}{2^k} > 2015$

Of course, $k = 2015$ won't work since the log is greater than zero. However $\boxed{k = 2016}$ does work because since $2015! > 2^{2016}$ (trivial), the log will be less than one.