Is $\infty^\infty$ indeterminate?

No, $\infty^\infty=\infty$. More formally:

If $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=\infty$, $\lim_{x\to a}f(x)^{g(x)}=\infty$.

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$$\forall x>\max\{N_f,\,N_g\}(f(x)^{g(x)}>N^N>M).$$

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 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.

If we denote $+\infty$ with $\infty$, then $\infty^\infty$ clearly diverges, but there are cases such as $${\infty^{-\infty}=0\\(-\infty)^\infty\in\{-\infty,\infty,\text{not exists}\}\\(-\infty)^{-\infty}\in\{0,\text{not exists}\}}$$