Note that $$\begin{aligned} \prod_{i=2}^{2012} \log _{i} \pi(i) & =\prod_{i=2}^{2012} \frac{\log \pi(i)}{\log i} \\ & =\frac{\prod_{i=2}^{2012} \log \pi(i)}{\prod_{i=2}^{2012} \log i} \\ & =1 \end{aligned}$$ where the last equality holds since $\pi$ is a permutation of the numbers 2 through 2012.