Exponential Factorial vs Tetration I'm wondering whether there's a known way to compare the exponential factorial of n versus the tetration of a fixed number $($ e.g., $3$, since it appears in Graham's number $)$ with the same number of terms.
Here's an example: Find the smallest n such that $n^{!}$ is greater than $^{n-1}3$.
Both of these terms have the same number of exponents $($we ignore $1$ in the factorial exponential, thus $n-1$ terms$)$. A first computation shows that the tetration of $3$ grows much faster:
$4^{!} = 262144,~$ while $~^{3}3 = 3^{27} \approx 7.62 \cdot 10^{12}$
$5^{!} = 5^{262144} \approx 10^{183230},~$ while $~^{4}3 = 3^{3^{27}} \approx 10^{3.6 \cdot 10^{12}}$
Afterwards, numerical evaluation becomes more difficult, but what appears to be clear is that until the basis for the exponent in the exponential factorial becomes much, much larger, the tetration of $3$ will keep growing insanely faster, simply due to the higher last exponent.
Is this a known problem ?
 A: In fact, for no $n$ do we have $n^!$ greater than $^{n-1} 3$.  For the following, we will use $\log$ to mean logarithm base 3.
We will prove by induction that $(n+2)^{n^!} < \ ^n 3$ for $n \ge 2$.
Base cases: $4^{2^1} = 16 < 27 = \ ^23$.
$5^{3^!} = 5^9 < 3^{27} = \ ^33$
$6^{4^!} = 6^{4^9} < 3^{2* 4^9} < 3^{3^{27}} = \ ^4 3$
Inductive step: Suppose that $(n+2)^{n^!} < \ ^n 3$ for some $n \ge 4$. Then
$\log\log ((n+3)^{(n+1)^!}) = \log((n+1)^! \log (n+3)) = \log((n+1)^!) + \log\log(n+3) = n^! \log(n+1) + \log \log (n+3)$.
Now, I claim that $n^! \log(n+1) + \log \log (n+3) < n^! \log(n+2)$.  After rearranging, this is equivalent to $n^! > \frac {\log\log(n+3)}{\log(n+2) - \log(n+1)}$.  But in fact
$\frac{\log\log(n+3)}{\log(n+2) - \log(n+1)} = \frac{\log\log(n+3)}{\log(1 + \frac{1}{n+1})} = \frac{\log\log(n+3)}{\frac{\ln(1+\frac{1}{n+1})}{\ln(3)}} < \frac{\log\log(n+3)}{\frac{1}{(n+2)\ln(3)}} = (n+2)\ln(3) \log\log(n+3) < n^9 \le n^!$.
(The last two inequalities follow because $n \ge 4$.)
So
$\log\log ((n+3)^{(n+1)^!}) <  n^! \log(n+2) = \log((n+2)^{n^!}) < \log(^n 3) = \ ^{n-1} 3$
$(n+3)^{(n+1)^!} < \ ^{n+1} 3$, as desired.
So for $n \ge 3$, $n^! < (n+1)^{(n-1)^!} < \ ^{n-1} 3$, and of course one can verify this is true for $n = 2$ as well.
Going in the other direction, obviously $n^! > \ ^{n-2} 3$ for $n \ge 2$, since each term of the first tower is greater or equal to each term of the second, with equality in just one case. So for all $n \ge 2$ we have
$^{n-2} 3 < n^! < \ ^{n-1} 3$.
