Comparing a Factorial and a Perfect Power Let us define the following recurrence relations as so. $$a_1=6, a_{n+1}=a_n!$$
$$b_1=6, b_{n+1}=6^{b_n}$$ 
So, which of the following is larger? $a_{b_2}$ or $b_{a_2}$?
To clarify, I am trying to compare $6^{6^{6^{6^6\dots}}}$ ($720$ times) and $(((6!)!)!)!\dots$($46656$ times) 
While I tried to use a calculator to determine which is larger, the values were too large for me to compute. 
Their log values proved difficult to compare as well. 
Wolframalpha did not prove much help(see here and here) 
Normally it is true that $a_n$ is far smaller then $b_n$, but because $b_2$ is far larger than $a_2$, I find it hard to determine. 
Any help would be appreciated. 
 A: Let $G(1)=6, G(n)=6^{G(n-1)}$ and $F(1)=6, F(n)=(F(n-1))!$.  You want to compare $G(720)$ with $F(46656)$.  We have $\log G(n)=G(n-1) \log (6)$, so $720$ applications of $\log$ makes it small.  $\log F(n)\approx F(n-1)( \log F(n-1)-1)$, so it takes about $46656$ applications of $\log$ to make it small.  $F(46656) \gg G(720)$ so much that dividing $F(46656)/G(720)$ does not make it appreciably smaller.
A: A proof by words(without math) would be -
 to get $b_{n+1}$ from $b_n$ , you multiply '$6$' $b_n$ times.
 to get $a_{n+1}$ from $a_n$ , you multiply number really higher than $6$ as $n$ increases. So, I would say, $a_n$ is greater than $b_n$ for large $n$.
As far as your question is concerned, $a_{b_2}$ is larger-
$a_{b_2}= (6^6 - 1)! = (6^6 - 1)*(6^6 - 2)....(6^6 - 120)*....$many numbers 
$b_{a_2}= (6^{6!})   = (6^6)*(6^6)....$$120$ times.
$6^6 - 120 \sim 6^6$ .
So, just by $120$ numbers in the factorial, you reach very close to $b_{a_2}$..
There are a lot more numbers after that.
From this, I would say, $a_{b_2} > b_{a_2}$
