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$\newcommand\iddots{\mathinner{ \kern1mu\raise1pt{.} \kern2mu\raise4pt{.} \kern2mu\raise7pt{\Rule{0pt}{7pt}{0pt}.} \kern1mu }}$What is the limit as $n$ tends to infinity of

$$\frac{2^{n^k}}{2^{2^{{\iddots}^2}}},$$

where the denominator is a tower of $n$ twos.

Seems like it should be zero, and I know that $\frac{n^k}{2^n}$ tends to zero.

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    $\begingroup$ To be clear, you mean that the denominator is tetration $2\uparrow\uparrow n$ and not $2^{2^{n-1}}$? $\endgroup$ – K B Dave May 16 at 0:02
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    $\begingroup$ Yep, correct: tower of 2s $\endgroup$ – user112358 May 16 at 0:03
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Since $n^k<<2^n$ then $\frac{2^{n^k}}{2^{2^n}}$ tends to $2^{-\infty}=0$.

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