$\underset{n\to \infty }{Lim}$ $\left\{{\log }_{(n-1)}\left(n\right){\log }_n\left(n+1\right)\mathrm{......}{\log }_{n^k-1}\left(n^k\right)\right\}$ is equal to

$\underset{n\to \infty }{Lt}[\frac{\cancel{\log n}}{\log \left(n-1\right)}.\frac{\cancel{\log \left(n+1\right)}}{\cancel{\log n}}.\frac{\log n^k}{\cancel{\log \left(n^k-1\right)}}]$ 

        $=\underset{n\to \infty }{Lt}\frac{k\log n}{\log \left(n-1\right)}$

        $=k\underset{x\to \infty }{Lt}\frac{\frac{1}{n}}{\frac{1}{n-1}}=k\left(\underset{n\to \infty }{Lt}\right)(1-\frac{1}{n})$

         = k