Which arrangement produces the largest number? I learnt that the power tower $2\uparrow3\uparrow4\uparrow...\uparrow n$ is larger
 than any power tower with a different order of the numbers $2,3,4,...,n$.
Is this also true for conway-chains and for bowers array notation ?
Are $$2\rightarrow 3\rightarrow 4\rightarrow...\rightarrow n$$
and {$2,3,4,...,n$} also larger than any other number of this form
 with different order of the numbers ?
 A: For Conway Chained Arrow Notation, the largest possible number is always:
$$3→2→4→...→n$$
For $n>4$ this follows directly from the recursion process given by Conway's notation plus the fact that $3^2>2^3$. For $n=4$, one must calculate it directly ($3↑↑↑↑2$ vs $2↑↑↑↑3$) to show that $3↑↑↑↑2$ is indeed larger.
As for Bowers Arrays, I'm not too familiar with them, but since $\lbrace{a,b,c}\rbrace=a→b→c$, then the largest 3-element array made with $2,3,4$ would be $\lbrace{3,2,4}\rbrace$}. 
A: No for both.
For example, for $n = 4$ we have $2 \rightarrow 3 \rightarrow 4 = 2 \rightarrow 4 \rightarrow 3 = 2 \uparrow \uparrow 65536$, whereas $3 \rightarrow 2 \rightarrow 4 = 3 \uparrow \uparrow 3^{27}$.
We can show by induction that $2 \rightarrow 3 \rightarrow n < 3 \rightarrow 2 \rightarrow n$, as
$$
2 \rightarrow 3 \rightarrow 1 = 8 < 9 = 3 \rightarrow 2 \rightarrow 1
$$
and assuming $2 \rightarrow 3 \rightarrow n = 2 \rightarrow 4 \rightarrow 
(n-1) < 3 \rightarrow 2 \rightarrow n = 3 \rightarrow 3 \rightarrow (n-1)$, we have
$$
2 \rightarrow 3 \rightarrow (n+1) = 2 \rightarrow (2 \rightarrow 2 \rightarrow (n+1)) \rightarrow n = 2 \rightarrow 4 \rightarrow n 
$$
$$
= 2 \rightarrow (2 \rightarrow 4 \rightarrow (n-1)) \rightarrow (n-1) < 2 \rightarrow (3 \rightarrow 3 \rightarrow (n-1)) \rightarrow (n-1)
$$
$$
< 3 \rightarrow (3 \rightarrow 3 \rightarrow (n-1)) \rightarrow (n-1) = 3 \rightarrow 3 \rightarrow n = 3\rightarrow 2 \rightarrow (n+1).
$$
It follows that $2 \rightarrow 3 \rightarrow X < 3 \rightarrow 2 \rightarrow X$ for any chain $X$, since when you evaluate the chains you will get the same expressions, except one will have chains starting with $3 \rightarrow 2$ and one will have chains starting with $2 \rightarrow 3$.  Whenever the expressions reduce a chain to a 3-chain, we will have $3 \rightarrow 2 \rightarrow n$ evaluate to a higher value than $2 \rightarrow 3 \rightarrow n$, so in the end $3 \rightarrow 2 \rightarrow X$ will evaluate to a higher value than $2 \rightarrow 3 \rightarrow X$.  In particular, $2 \rightarrow 3 \rightarrow \cdots \rightarrow n < 3 \rightarrow 2 \rightarrow \cdots \rightarrow n$.
For Bowers arrays the situation is more extreme.  Any expression of the form {2,2,...} will evaluate to the number 4, as applying the evaluation rules will either keep the first two entries the same, or will replace the array with {2, {2,1,...},...} = {2,2,...}. So the array will eventually evaluate to {2,2} = 4.
Further, any array {2,b,c,d,...} with four or more entries (not counting trailing 1's) will evaluate to 4 as well, as {2,b,c,d,...} will evaluate to {2,b',c-1,d,...} and then {2,b'',c-2,d,...} eventually reaching {2,n,1,d,...}. This evaluates to {2,2,{2,n-1,1,d,...},d-1,...}, which we have already determined equals 4.
In particular, {2,3,4,...,n} will evaluate to 4 for n > 4, whereas {3,2,4,...,n} will grow extremely fast.
