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Tetration is repeated exponentiation evaluated from right to left. The value of both the factorial and tetration function at $x=0$ is defined to be 1. So,both functions start at $x=0$ (when the values of both of them is 1) and by overtaking $^{x}10$, I mean approaching infinity faster than tetration as $x$ is increased.

$x!$ is definitely slower. I've no idea how large $x!!$ can become but even for $x=3$ the value of $^310$ would mean $10000000000$ zeros on 1. So, I don't think the double factorial grows faster either. Can any number of factorials applied to $x$ be faster?

As a side note, I think this function overtakes tetration in the same way factorials overtake exponentiation: $$f(x)=x^{(x-1)^{(x-2)^{.........^{2^1}}}}$$

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  • $\begingroup$ @BrevanEllefsen: Don't you need ${}^x n$ rather than ${}^n x$ for that to be relevent? $\endgroup$ – hmakholm left over Monica Mar 7 '17 at 15:38
  • $\begingroup$ See D. FACTORIALS AND EXPONENTIATION OF TETRATED NUMBERS here. For all but very small values of $n$ (say, $n < 4),$ applying any of the functions $2^x,$ $10^x,$ $x!,$ $x^x$ to $^{x}10$ essentially produces $^{x+1}10,$ and so applying any of these functions $n$ many times to $^{x}10$ essentially produces $^{x+n}10.$ I believe applying your function $f(x)$ to $^{x}10$ produces something less than $^{2x-1}10.$ Thus, tetration is so far beyond these other functions that applying them to tetrated numbers doesn't tetrate-affect them much. $\endgroup$ – Dave L. Renfro Mar 7 '17 at 15:39
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We know that $n! \lt n^n=~^2n$ from Stirling's approximation. Then $n!! \lt (n^n)^{n^n}=n^{(n^{n+1})}$ which is not so different from $^3n$. $~n!!! \lt (n^{(n^{n+1})})^{(n^{(n^{n+1})})}=n^{(n^{n+1})(n^{(n^{n+1})})}\approx~^4n$ The important thing in tetration is the height of the tower-the numbers that are in it are much less important. Each factorial will essentially add one layer because all the powers in the lower level get absorbed into the top level, so $k$ applications of factorial should be very roughly comparable to $^{k+1}n$

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