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$$\prod_{x=2}^\infty\frac x{(\ln x)^x}$$ 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?

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    Note that $x^y=e^{y\cdot ln(x)}$. Thus your denominator $ln^x(x)$ is equal to $e^{x\cdot ln(ln(x))}$. Yeah, it grows quite fast. :) – 2017-01-10

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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.