What can be said about the value of $\infty^\infty$?
Some of the main arguments regarding indeterminations like $0^0$ and $1^\infty$ involve the fact that different limit directions yield different results, in some cases only noticable via the complex plane. Are there similar arguments that can be used for this case, or is it simply $\infty$?
On
No. The only common indeterminate forms are $0\over0$, $\infty\over\infty$, $0\cdot\infty$, $1^\infty$, $\infty-\infty$, $0^0$, and $\infty^0$. As J.G. notes, $\infty^\infty$ is clearly unbounded in the positive direction.
On
If we denote $+\infty$ with $\infty$, then $\infty^\infty$ clearly diverges, but there are cases such as $${\infty^{-\infty}=0\\(-\infty)^\infty\in\{-\infty,\infty,\text{not exists}\}\\(-\infty)^{-\infty}\in\{0,\text{not exists}\}}$$
On
No. Its undefined. There is no ambiguity in what value it might converge to, if it did so. For example $0^0$ could be 0 or 1, depending on interpretation. Its neither, in fact, what we call indeterminate because there is no reliable way of treating the problem. The same cannot be said for $\infty^\infty$. Infinity is not a number, which is why we call it undefined. Indeterminate is reserved for expressions for which we cannot determine a singular correct answer; in fact expressions like $0/0$ or $0^0$ could be made to equal anything which is why they equal nothing.
On
First, it is worth commenting on the precision of mathematical language. Neither $0^0$ nor $1^\infty$ are indeterminate forms. Rather, these are mathematical expressions which are generally left undefined. When we discuss indeterminate forms, what we are discussing are expressions which, when we take a limit by naively applying the properties of continuity, fail to result in meaningful expressions. For example: $$\lim_{x\to\infty} \left( 1 + \frac{1}{x} \right)^{x} \qquad\text{vs}\qquad \lim_{x\to\infty} \left( 1 - \frac{1}{x} \right)^{x}. $$ In either case, if we try to naively compute, we get $$ \lim_{x\to\infty} \left( 1 \pm \frac{1}{x} \right)^{x} = \left[\lim_{x\to\infty} \left( 1 \pm \frac{1}{x} \right) \right]^{\lim_{x\to\infty} x} "=" 1^{\infty} $$ (note that I have put the last equality in quotes—it is not true that the expression on the left is equal to $1^{\infty}$, since $1^{\infty}$ is not even well-defined). Because $1^{\infty}$ does not really make sense as a mathematical expression on its own we say that the given limits are "indeterminate of the type $1^{\infty}$." It is the limit which is of indeterminate form, not the expression obtained by naive computation of the limit.
Note in the example above that $$ \lim_{x\to\infty} \left( 1 + \frac{1}{x} \right)^{x} = \mathrm{e} \qquad\text{while}\qquad \lim_{x\to\infty} \left( 1 - \frac{1}{x} \right)^{x} = \frac{1}{\mathrm{e}}. $$ The two limits are of same form (as noted above, if we attempt to compute them using "limit laws", we get the meaningless expression $1^{\infty}$), but they are not equal to each other. Thus their form is indeterminate.
With the above in mind, the original question can probably be rephrased as
Question: Are there indeterminate limits which are of the form $\infty^{\infty}$?
In other words, if we have two functions $f$ and $g$ such that $$ \lim_{x\to a} f(x) = +\infty \qquad\text{and}\qquad \lim_{x\to a} g(x) = +\infty, $$ can we definitely determine the value of $$ \lim_{x\to a} \left( f(x)^{g(x)} \right), $$ or might this limit depend on the specific functions $f$ and $g$?
This version of the question is answered very nicely by J.G., and I will add nothing to their arguments, other than to provide a short answer to the version of the question I gave above:
Answer: No, there are no indeterminate limits of the form $\infty^{\infty}$. If a limit is of the form $\infty^\infty$, then that limit is itself infinite.
No, $\infty^\infty=\infty$. More formally:
I'll prove the case $a=\infty$; you can do others as an exercise. Firstly, though, note that theorems such as this are how we define $b^c$ when we're not sure what it is, in the way you're sure what $2^5$ is. There is no similar theorem to define $1^\infty$, as can be seen by considering the behaviour of various $(\exp 1/x)^{cx}$ for $c\in\Bbb R$.
Now to the proof. For large positive $M$, choose positive $N$ large enough that $\forall x>N(x^x>M)$, then choose positive $N_f,\,N_g$ large enough that $\forall x>N_f(f(x)>N)$, and similarly for $g$'s divergence. Then$$\forall x>\max\{N_f,\,N_g\}(f(x)^{g(x)}>N^N>M).$$