Graham's number G and the Moser number M are both humungously large, but G is very much larger than M.
In any expression involving Knuth arrows and positive integers n > 2, the parenthisation that creates the largest number is the one that associates from the right.
For example, (a^^^b)^^c < a^^^(b^^c)
Proof: by induction.
For single arrows, or ordinary exponentiation, this is the well known result that (a^b)^c < a^(b^c), or that a^(b×c) < a^(b^c), or that b×c < b^c, which is true for b, c > 2.
If n>2, then n[k+2] < n^^...^^n (2k-1 arrows)
Proof. From the details of the Moser construction, it is easy to see that n[4] = (...(n^n)^(n^n)...)^...^(...(n^n)^(n^n)...)[2n terms]. From Lemma 1, this is less than the expression with n terms all associating to the right: n[4] < n^n^...^n[2^n terms]. From the definition of Knuth arrows, we have n[4] < n^^2^n
It is not hard to show, for n>2, that n^^2^n <n^^^n
Then one can proceed to the general result by induction on k, invoking Lemma 1 in the general case.
For example, when k=3 (and n>2):
n[5] = n[4]n [from definition]
= n[4][4]n-1
< (n^^^n)[4]n-1 [by Lemma 2, with k = 2]
< ((n^^^n)^^^(n^^^n))[4]n-2 [by Lemma 2 again]
...
< (...(n^^^n)^^^(n^^^n)...)^^^...^^^(...(n^^^n)^^^(n^^^n)...) (2n terms)
< n^^^^2^n [by Lemma 1]
< n^^^^^n
M = 2[2[5]] < 3[2[5]] < 3^^...^^3 (2[5]×2 -1 arrows), by lemma 2.
Now 2[5] < 3[5] < 3^^^^^3 by lemma 2
So M < 3^^...^^3 (3^^^^^3×2-1 arrows) << G2 << G
[7 July 1998. My thanks to Tim Chow for emailing me this proof.]