Is there a way to prove or define or delineate all the indeterminate forms? Is there a way to know what any given indeterminate form looks like? Can we prove there are only so many finite cases of them?
On Wikipedia I see $\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$, $0 \cdot \infty$, $1^{\infty}$, $\infty - \infty$, $0^0$, and $\infty^0$, but are there any others? Can we prove that these are indeterminate? Are there finitely many? Infinitely many? 
 A: Using the fact that for $x>0$ and $\alpha\in \Bbb R $, we have
$$x^\alpha=e^{{\alpha}\ln (x)}. $$
we deduct that your indeterminate form $(+\infty)^0$ is the same than $0.(+\infty) $.
A: The indetrminate forms are precisely those forms $a\circ b$ with $a,b$ from the extendeed reals (i.e., including infinities) and $\circ$ a binary operation that has been extended to the extended reals, where the map $\langle x,y\rangle \mapsto x\circ y$ is not continuous or cannot be extended continuously.
In other words, $a_n\to a$ and $b_n\to b$ does not necessarily imply $a_n\circ b_n\to a\circ b$.
So for example $\frac 00$ is an indetermined form because the division, which (whithout considering infinite inputs) is defined on $\Bbb R\times (\Bbb R\setminus\{0\}$) and cannot be extended continuously to $\langle 0,0\rangle$.
Likewise $0^0$ is an indeterminate form because, while $0^0=1$ is defined, exponentiation is not continuous there.
Likewise $\infty-\infty$ is an indeterminate form because, even if you should boldly define arithmetic on the extended reals to have $\infty-\infty=0$, subtraction is not continuous at infinity.
Now note that the standard arithmetic operations $+,-,\cdot,/$ are "mostly harmless", that is, they are continuous where they are defined in real numbers. Thus their indeterminate forms can occur at most where they are not defined (which is only division by $0$) and/or where we attempt to play with infinite inputs. Exponentiation is a bit more complex because there are competing definitions and not a single definition that covers all cases; "most" cases are covered by $x^y:=\exp(y\ln x)$, which is of course continuous where defined. So exceptional cases occur with infinite input or when $x$ is not positive.
The preceding paragraph seems to suggest that more than the well-known indeterminate forms are possible, e.g., $2^{+\infty}$, but those are only candidates for additional indeterminate forms. A closer look immediatly shows that $a_n\to 2$ and $b_n\to +\infty$ unambiguously implies $a_n^{b_n}\to +\infty$.
