Here we are going to prove that 100^^^100 < E3#100#99 < 10^^^101 < E100#100#100 < E1#102#100 < 10^^^102, to establish the validity of this order on my ULNL. We'll begin with the smallest value and work our way up.
Firstly we have:
100^^^100 = 100^^100^^100^^ ... ^^100^^100 w/100 100s
What we want to do first is be able to change the base 100s to 10s. To do this we'll begin with tetration. Observe:
100 < 10^3 = E3#1
100^100 = 10^(2*100) = 10^(2*10^2) < 10^10^3 = E3#2
100^^3 = 100^100^100 = 10^(2*10^(2*10^2)) < 10^(10*10^(2*10^2)) = 10^10^(1+2*10^2) =
10^10^201 < 10^10^1000 = 10^10^10^3 = E3#3
100^^4 = 100^100^^3 < 10^(2*100^^3) < 10^(10*100^^3) = 10^(10*100^100^^2) =
10^(10*10^(2*100^^2)) = 10^10^(2*100^^2+1) < 10^10^(10*100^^2) = 10^10^(10*10^(2*10^2)) =
10^10^10^(2*10^2+1) < 10^10^10^10^3 = E3#4
As you can see 100^^n never makes it to E3#n. This can be proven by a method I'll call "recurrence".
Say we have an expression of the form:
10^10^ ...^10^(2*100^^k+1)
where k is any positive integer, and k>1.
We can bound the top most term as of the power tower as follows:
2*100^^k+1 < 3*100^^k = 3*100^100^^(k-1) < 10*100^100^^(k-1) =
10*10^(2*100^^(k-1)) = 10^(2*100^^(k-1)+1)
So we have a recurrence transformation 2*100^^k+1 --> 10^(2*100^^(k-1)+1). Note that this is true for any k>1.
Now observe how we can use this transformation to prove that 100^^n must be less than E3#n:
If n=1, then 100^^1 = 100 = 10^2 < 10^3 = E3#1
If n=2
100^^2 = 100^(100^^1) = 10^(2*100^^1) = 10^(2*100) = 10^(2*10^2) = 10^200 < 10^1000 = 10^10^3 = E3#2
If n>2
100^^n = 100^100^^(n-1) = 10^(2*100^^(n-1)) = 10^(2*100^(100^^(n-2))) =
10^(2*10^(2*100^^(n-2))) < 10^(10*10^(2*100^^(n-2))) =
10^10^(2*100^^(n-2)+1)
At this point we can apply recurrence until the tetrate is reduced to 1:
10^10^(2*100^^(n-2)+1) < 10^10^(10^(2*100^^(n-3)+1)) < 10^10^(10^(10^(2*100^^(n-4)+1)))
< ... ... ... ... < 10^10^(10^(10^ ... (10^(10^(2*100^^1+1))) ... )) w/n-1 10s =
E(2*100^^1+1)#(n-1) = E201#(n-1) < E1000#(n-1) = E10^3#(n-1) = E3#n.
Thus we conclude:
100^^n < E3#n : n=Z+ [Lemma 1, L1]
With this we now show that 100^^^100 < E3#100#99. Simply observe:
100^^100 < E3#100 [L1]
100^^^3 = 100^^100^^100 < 100^^E3#100 [L1] < E3#(E3#100) [L1] = E3#100#2
100^^^4 = 100^^100^^^3 < 100^^E3#100#2 < E3#(E3#100#2) = E3#100#3
...
100^^^n < E3#100#(n-1)
:: 100^^^100 < E3#100#99
This is already enough to show that 100^^^100 is less than a greagol, since 100^^^100 < E3#100#99 < E100#100#100 = greagol.
To prove that E3#100#99 < E1#1#101 I'll use a method called the "ascending one". The basic idea of the ascending one is that if you have any positive integer power of ten, then:
10^n + 1 < 10^(n+1) : n=Z+
This is simple enough to see. Since 1 < 10^1 < 10^2 < 10^3 < ... etc. It follows that:
10^n + 1 < 10^n + 10^n = 2*10^n < 10*10^n = 10^(n+1)
The one has "ascended". This leads to a nice an simple rule of thumb that says, if a one is moved from a lower position of a power tower to a higher one, then the new value will be much greater. It should be noted that this can be generalized in the following way:
10^n + m < 10^(n+1) : n=Z+ , whenever m =< 10^n
We'll now apply the ascending method to our problem. Remember that Ea#b = 10^10^...^10^a w/b 10s. So 1+Ea#b < 10^(1+Ea#(b-1)) < 10^10^(1+Ea#(b-2)) < ... < 10^10^...^10^(1+a) = E(a+1)#b.
So we can say:
1+Ea#b << E(a+1)#b : a,b=Z+ [Ascended one, or AO]
Also important to note is that:
E10#b = E1#(b+1)
This last statement will be very useful in the following proof:
E3#100#99 = E3#(E3#100#98) < E10#(E3#100#98) =
E1#(1+E3#100#98) < E1#(1+E3#(E3#100#97)) < E1#(E4#(E3#100#97)) [AO] <
E1#(E10#(E3#100#97)) = E1#(E1#(1+E3#100#97)) < E1#(E1#(E4#(E3#100#96)))
< ... recurrence ... < E1#(E1#(E1#( ... (E1#(E1#(E4#(E3#100#1)))) ... ))) w/97 E1#s =
E1#(E4#(E3#100#1))#97 < E1#(E10#(E3#100#1))#97 = E1#(E1#(1+E3#100#1))#97 =
E1#(E1#(1+E3#100))#97 < E1#(E1#(E4#100))#97 [AO] = E1#(E4#100)#98 <
E1#(E10#100)#98 = E1#(E1#101)#98 = E1#101#99 < E1#10^^10#99 = E1#1#101
Not bad! So we've established that 100^^^100 < E3#100#99 < E1#1#101.
Next we can show that E1#1#101 is less than greagol as follows:
E1#1#101 = E1#10#100 < E100#100#100 = greagol.
To show that greagol < E1#102#100, we have to go back again to the ascending method:
E100#100#100 = E100#(E100#100#99) < E10^10#(E100#100#99) = E1#(2+E100#100#99) <
E1#(E101#(E100#100#98)) [AO] < E1#(E1#(2+E100#100#98)) < E1#(E1#(E101#(E100#100#97))) =
E1#(E101#(E100#100#97))#2 < E1#(E101#(E100#100#96))#3 < ... recurrence ... <
E1#(E101#(E100#100#1))#98 < E1#(E10^10#(E100#100))#98 = E1#(E1#(2+E100#100))#98 =
E1#(2+E100#100)#99 < E1#(E101#100)#99 < E1#(E1#102)#99 =
E1#102#100
So greagol < E1#102#100.
The last part is quite easy to show:
E1#102#100 < E1#10^^10#100 = E1#1#102 = 10^^^102.
Thus we have established the order:
10^^^100 < 100^^^100 < E3#100#99 < 10^^^101 < E100#100#100 < E1#102#100 < 10^^^102