$$\prod_{x=2}^\infty\frac x{(\ln x)^x}$$ I tried checking the answer at Wolfram Alpha, and it gave me the answer $≈0.×10^{-100}$.
Does the denominator of $(\ln x)^x$ make it so large that the product tends to zero?
How to actually evaluate it, or at least how to find a nice approximation?
For $x>e^e$ you have $\log x>e$ and $\log^x x>e^x$ so the sequence $\frac{x}{\log^x x}\to 0$, and thus the product is zero.