Popular mathematics folklore provides some simple tools
enabling us compactly to describe some truly enormous
numbers. For example, the number $10^{100}$ is commonly
known as a googol,
and a googol
plex is
$10^{10^{100}}$. For any number $x$, we have the common
vernacular:
 $x$ bang is the factorial number $x!$
 $x$ plex is the exponential number $10^x$
 $x$ stack is the number obtained by iterated exponentiation
(associated upwards) in a tower of height $x$, also denoted $10\uparrow\uparrow x$,
$$10\uparrow\uparrow x = 10^{10^{10^{\cdot^{\cdot^{10}}}}}{\large\rbrace} x\text{ times}.$$Thus, a googol bang is $(10^{100})!$, and a googol stack is
$10\uparrow\uparrow 10^{100}$. The vocabulary enables us to
name larger numbers with ease:
 googol bang plex stack. (This is the exponential tower $10^{10^{\cdot^{\cdot^{^{10}}}}}$ of height $10^{(10^{100})!}$)
 googol stack bang stack bang
 googol bang bang stack plex stack
 and so on…
Consider the collection of all numbers that can be named in
this scheme, by a term starting with googol and having
finitely many adjectival operands: bang, stack, plex, in
any finite pattern, repetitions allowed. (For the purposes
of this question, let us limit ourselves to these three
operations and please accept the base 10 presumption of the
stack and plex terminology simply as an artifact of its
origin in popular mathematics.)My goal is to sort all such numbers nameable in this
vocabulary by size.A few simple observations get us started. Once $x$ is large
enough (about 20), then the factors of $x!$ above $10$
compensate for the few below $10$, and so we see that
$10^x\lt x!$, or in other words, $x$ plex is less than $x$
bang. Similarly, $10^{10^{:^{10}}}x$ times is much larger
than $x!$, since $10^y\gt (y+1)y$ for large $y$, and so for
large values we have
 $x$ plex $\lt$ $x$ bang $\lt$ $x$ stack.
In particular, the order for names having at most one
adjective is:googol googol plex googol bang googol stack
And more generally, replacing plex with bang or bang with
stack in any of our names results in a strictly (and much)
larger number.Continuing, since $x$ stack plex $= (x+1)$ stack, it
follows that
 $x$ stack plex $\lt x$ plex stack.
Similarly, for large values,
 $x$ plex bang $\lt x$ bang plex,
because $(10^x)!\lt (10^x)^{10^x}=10^{x10^x}\lt 10^{x!}$.
Also,
 $x$ stack bang $\lt x$ plex stack $\lt x$ bang stack,
because $(10\uparrow\uparrow x)!\lt (10\uparrow\uparrow
x)^{10\uparrow\uparrow x}\lt 10\uparrow\uparrow 2x\lt
10\uparrow\uparrow 10^x\lt 10\uparrow\uparrow x!$. It also
appears to be true for large values that
 $x$ bang bang $\lt x$ stack.
Indeed, one may subsume many more iterations of plex and
bang into a single stack. Note also for large values that
 $x$ bang $\lt x$ plex plex
since $x!\lt x^x$, and this is seen to be less than
$10^{10^x}$ by taking logarithms.The observations above enable us to form the following
order of all names using at most two adjectives.googol googol plex googol bang googol plex plex googol plex bang googol bang plex googol bang bang googol stack googol stack plex googol stack bang googol plex stack googol bang stack googol stack stack
My request is for any or all of the following:
Expand the list above to include numbers named using
more than two adjectives. (This will not be an
endextension of the current list, since googol plex plex
plex and googol bang bang bang will still appear before
googol stack.) If people post partial progress, we can
assemble them into a master list later.Provide general comparison criteria that will assist
such an ongoing effort.Provide a complete comparison algorithm that works for
any two expressions having the same number of adjectives.Provide a complete comparison algorithm that compares
any two expressions.Of course, there is in principle a computable comparison
procedure, since we may program a Turing machine to
actually compute the two values and compare their size.
What is desired, however, is a simple, feasible algorithm.
For example, it would seem that we could hope for an
algorithm that would compare any two names in polynomial
time of the length of the names.
Answer
OK, let’s attempt a sorting of the names having at most
three operands. I’ll make several observations, and then
use them to assemble the order section by section,
beginning with the part below googol stack.

googol bang bang bang $\lt$ googol
stack. It seems clear that we shall be able to iterated bangs many
times before exceeding googol stack. Since googol bang bang
bang is the largest threeoperand name
using only plex and bang, this means that all such names will interact
only with each below googol stack. 
plex $\lt$ bang. This was established in the question.

plex bang $\lt$ bang plex. This was established in the
question, and it allows us to make many comparisons in
terms involving only plex and bang, but not quite all of
them. 
googol bang bang $\lt$ googol plex plex plex. This is
because $g!!\lt (g^g)^{g^g}=g^{gg^g}=10^{100\cdot gg^g}$, which is less than
$10^{10^{10^g}}$, since $100\cdot gg^g=10^{102\cdot
10^{100}}\lt 10^{10^g}$. Since googol bang bang is the largest twooperand name using only
plex and bang and googol plex plex plex is the smallest threeoperand name, this means that
the twooperand names using only plex and bang will all come
before all the threeoperand names. 
googol plex bang bang $\lt$ googol bang plex plex. This
is because $(10^g)!!\lt
((10^g)^{10^g})!=(10^{g10^g})!=(10^{10^{g+100}})!\lt
(10^{10^{g+100}})^{10^{10^{g+100}}}=10^{10^{g+100}10^{10^{g+100}}}=
10^{10^{(g+100)10^{g+100}}}\lt
10^{10^{g!}}$.
Combining the previous observations leads to the following
order of the threeoperand names below googol stack:
googol
googol plex
googol bang
googol plex plex
googol plex bang
googol bang plex
googol bang bang
googol bang bang
googol plex plex plex
googol plex plex bang
googol plex bang plex
googol plex bang bang
googol bang plex plex
googol bang plex bang
googol bang bang plex
googol bang bang bang
googol stack
Perhaps someone can generalize the methods into a general
comparison algorithm for larger smallish terms using only
plex and bang? This is related to the topic of the Velleman
article linked to by J. M. in the comments.
Meanwhile, let us now turn to the interaction with stack.
Using the observations of the twooperand case in the
question, we may continue as follows:
googol stack plex
googol stack bang
googol stack plex plex
googol stack plex bang
googol stack bang plex
googol stack bang bang
Now we use the following fact:
 stack bang bang $\lt$ plex stack. This is established as
in the question, since $(10\uparrow\uparrow x)!!\lt
(10\uparrow\uparrow x)^{10\uparrow\uparrow x}!\lt$
$(10\uparrow\uparrow x)^{(10\uparrow\uparrow
x)(10\uparrow\uparrow x)^{10\uparrow\uparrow x}}=$
$(10\uparrow\uparrow x)^{(10\uparrow\uparrow
x)^{1+10\uparrow\uparrow x}} 10\uparrow\uparrow 4x\lt
10\uparrow\uparrow 10^x$. In fact, it seems that we will be
able to absorb many more iterated bangs after stack into
plex stack.
The order therefore continues with:
googol plex stack
googol plex stack plex
googol plex stack bang
 plex stack bang $\lt$ bang stack. To see this, observe
that $(10\uparrow\uparrow 10^x)!\lt (10\uparrow\uparrow
10^x)^{10\uparrow\uparrow 10^x}\lt 10\uparrow\uparrow
2\cdot10^x$, since associating upwards is greater, and this
is less than $10\uparrow\uparrow x!$. Again, we will be
able to absorb many operands after plex stack into bang
stack.
The order therefore continues with:
googol bang stack
googol bang stack plex
googol bang stack bang
 bang stack bang $\lt$ plex plex stack.
This is because $(10\uparrow\uparrow x!)!\lt
(10\uparrow\uparrow x!)^{10\uparrow\uparrow x!}\lt
10\uparrow\uparrow 2x!\lt 10\uparrow 10^{10^x}$.
Thus, the order continues with:
googol plex plex stack
googol plex bang stack
googol bang plex stack
googol bang bang stack
This last item is clearly less than googol stack stack, and
so, using all the pairwise operations we already know, we
continue with:
googol stack stack
googol stack stack plex
googol stack stack bang
googol stack plex stack
googol stack bang stack
googol plex stack stack
googol bang stack stack
googol stack stack stack
Which seems to complete the list for threeoperand names.
If I have made any mistakes, please comment below.
Meanwhile, this answer is just partial progress, since we
have the fouroperand names, which will fit into the
hierarchy, and I don’t think the observations above are
fully sufficient for the fouroperand comparisons, although
many of them will now be settled by these criteria. And of course, I am nowhere near a general comparison algorithm.
Sorry for the length of this answer. Please post comments if I’ve made any errors.
Attribution
Source : Link , Question Author : JDH , Answer Author : JDH