I have to say I disagree with you.

Consider sequence A1=a, An+1=a^An for a>0.  If the limit x exists, we can take the limit on both sides and will get x=a^x, which is lna=ln(x)/x.  For a is in (0,1), the limit is the solution of lna=ln(x)/x, which is less than 1.

ln(x)/x is 0 at x=1 and x->infinity.  The max value occurs at e^(1/e). 

When a is in (1, e^(1/e)),
lna=ln(x)/x has two solutions.  We need to use the one that is less than e^(1/e) as the limit.

An example is a=2^.5.  (1)An is increase. (2)An<2

Proof of (2)

A1=a<2

Assume An<2 then An+1=a^An <a^2=2.

Therefore An<2 for all n.

Since (1)An is increase. (2)An<2, An has a limit.  Solve ln(2^.5)=ln(x)/x, we get x=2 or x=4. Accept 2 and reject 4 since it is bigger than e^(1/e).