So a friend started, in boredom, calculating values of what I have formalized as $f_n(a) = a^{f_{n-1}(a)}$ (also $f_0(a)=a$) for $a = 1.1$. He noticed that on his calculator it was not changing value from $f(n)$ to $f(n+1)$. I explained that it is increasing just the change is too small for floating point numbers so you just get the same values over and over again. Now we wan't to know what this is approaching.
It seems like it is changing by less and less for each n-1, n which indicates to me that there must be some limit. In other words it seems to me that $\lim_{n\rightarrow \infty} f_n(1.1) - f_{n-1}(1.1)$ is zero. This indicates that there is some finite limit of this thing.
So what is $\lim_{n\rightarrow \infty} f_n(1.1)$. My friend noticed that 1.4 has similar behavior but 1.5 does not. What about for other $a$? Certainly this only works for small $a$
Define $f(x)=a^x$. Note that your function $f_n$ is just applying $f$ to itself $n$ times. Now suppose that $\lim_{n\to\infty}f_n(a)$ exists, and call this limit $L$. Because the function $f$ is continuous, we know that $f(L)=L$:
\[f(L)=f(\lim_{n\to\infty}f_n(a))=\lim_{n\to\infty} f(f_n(a))=\lim_{n\to\infty} f_{n+1}(a)=L.\]
So the question becomes, for which values of $a$ can there exist some $L$ such that $a^L=L$ In other words, which values of $a$ can be expressed as $L^{1/L}$ for some $L$?
Now the problem simply becomes determining the range of the function $g(x)=x^{1/x}$. A quick plot of this function should convince you that it can attain any positive value less than a certain maximum. To determine that maximum, we set the derivative equal to zero.
\[0 = g'(x) = x^{-2+1/x}\cdot(1-\log(x))\]
For this to be zero, it must be that $1-\log(x)=0$, and so the maximum is attained at $x=e$. Thus, the maximum is $g(e)=e^{1/e}\approx1.44467$.
In other words, if you have $1<a\le e^{1/e}$, then a limit will exist. If $a>e^{1/e}$, no limit will exist. This explains why you found different behavior at $a=1.4$ and $a=1.5$. To deal with $a\le 1$ is a separate case I won't deal with here.
EDIT: I didn't explain why the case $a<1$ is different from the case $a\ge1$. The point is that the iterations will be increasing when $a>1$, which, along with the fact that they are bounded, is enough to show that the limit really exists. When $a<1$, all of the iterations are bounded, but the values bounce back and forth. Sometimes they will converge, but not always. Plot the points for $a=0.04$ and you'll see what I mean.