Well done Paul Jefferys, you got close to a complete solution here. We have to consider two different values of these climbing powers depending on the order of operations which can be shown by putting in brackets.

We can define 234 either as (23)4 = 212 or as 2(34) = 281.

In the same way there are two interpretations of √2√2√2

The first of these is f(f(√2)) where f(x)=x√2 which gives:

(√2√2)√2 = √2√2 ×√2 = √22 = 2

In the second case we get g(g(√2)) where g(x) = (√2)x, and using a calculator to get an approximate value gives:

√2(√2√2) = √2 1.63... = 1.76  to 2 decimal places

So

√2(√2√2) < (√2√2)√2.

Now consider

√2√2√2√2√2√2..
where the powers of √2 go on forever. We have seen that we have two possibilities, namely
X1 = lim
xn

where x1 = √2, xn+1 = xn√2 or
X2 = lim
xn

where x1 = √2, xn+1 = (√2)xn.

N.B. Both iterations can be done on a calculator or computer:
X1 = lim
xn

is equivalent to iterating f(x) = x√2 and
X2 = lim
xn

is equivalent to iterating g(x) = (√2)x. If you do this experimentally, in each case starting with x1=√2, you will find that the first iteration appears to converge to infinity and the second appears to converge to 2.

We claim X1 = +∞.

Proof
We have x1=√2 and xn+1=xn√2; thus

logxn+1=√2 logxn,     logx1 = log√2.

Thus


logxn+1 =
√2
 n
 
 log√2,

and as logxn → +∞ as n→ ∞, we see that xn→ +∞.

We now claim that X2=2.

First we show that xn < 2 for all n, and the proof is by induction. Clearly x1 < 2. Now suppose that xn < 2 and consider xn+1. We have


xn+1 =
√2
 xn
 
 <
√2
 2
 
 = 2

as required. Hence (by induction) xn < 2 for all n.

Next, we show by induction that xn < xn+1. It is clear that

x1 = √2 < (√2)√2 = x2.

Now suppose that xn−1 < xn. Then


 xn+1
xn
 =  √2xn
√2xn−1
 =√2xn−xn−1.

Thus, as xn−xn−1 > 0, we have xn+1/xn > 1 and hence xn+1 > xn. This shows that xn is increasing with n, and that xn < 2, and this is enough to see that xn converges to some number X2, where X2 ≤ 2. As xn+1=(√2)xn, if we let n tend to infinity we see that X2 is a solution of the equation x=(√2)x. If we now plot the graphs of y=x and y=(√2)x, we see that these two graphs meet at only two points, namely (2,2) and (4,4). Thus X2 is either 2 or 4, and so it must be 2.