Keep Logging....!

Using the change-of-base rule, the sum becomes

$log_{n}(2) + log_{n}(3) + log_{n}(4) + ... + log_{n}(2016)$ .

By the properties of logarithms we know that this then equals

$log_{n}(2 * 3 * 4 * ..... * 2016) = log_{n}(2016!)$ .

So with $n = 2016!$ we have the solution $log_{n}(2016!) = \boxed{1}$ .