The n-th of what?

NOTE: This is my first time using LaTeX (or writing an answer), have mercy on me for the spacing errors. Also, I'm pretty sure none of the answer choices are right. $\text{First, some algebraic manipulation:} \\ \log { \prod _{ i=1 } x_{ i } } =\log { x_{ 1 } } +\log { x_{ 2 } } +\dots \log { x_{ i } }$ $\\ \text{Therefore we now have:} \\ \ln { \prod _{ n=1 }^{ \infty }{ \sqrt [ n^{ x } ]{ \frac { 1 }{ n } } } } =\sum _{ n=1 }^{ \infty }{ \frac { -\ln { (n) } }{ n^{ x } } }$

$\\ \text{In the integral form below, a good technique that comes to mind is the application of Fubini's theorem. For this, we need the to make sure that neither} \\ \text {the sum nor the integral explodes (precisely, they must converge absolutely, but the integrand is strictly positive). I will only check the sum, and we} \\ \text{will see that the integral converges after interchanging the sum and the integral. }\\ \int _{ 2 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { \ln { (n) } }{ n^{ x } } } } dx$ $\\ \text{We observe the following in the limit of large n. Since this only gets "worse" in higher powers, it is sufficent to check if the sum converges for x=2.} \\ \ln { (n) } \ll n^{ 2 }$ $\\ \text{The final answerline is the left to the reader as an integration by parts problem, but shows that the integral converges.} \\\int _{ 1 }^{ \infty } \frac { \ln { (n) } }{ n^{ 2 } } dx =\int_0^{\infty}{ue^{-u}}dx$

$\\ \text{Now interchanging...} \\ \sum_{n=1}^{\infty}\ln{(n)}\int_2^{\infty}n^{-x}dx$ $\\ \text{Where x is evaluated at 2 and infinity. (I couldn't get the "bar" to format correctly)}\\ \sum_{n=1}^{\infty} -e^{-x\ln{(n)}}$
$\\ \text{Surprisingly, none of the answers seem to be right! The difference can be explained if the sum was taken from 2 to infinity and if the writer (or me)} \\ \text{missed a sign. Regardless, the closest answer (and [possibly] correct) contains zeta. }\\ \sum_{n=1}^{\infty}\frac{1}{n^2}= \zeta{(2)} \\ \text{I may have missed something, reply if you found an error! Thanks all.}$