No, because there's no such thing. Either we can define 0^0 to have a value, or we can not do so. Either way, there's no separate option "indeterminate" that is different from "undefined".
Now, it's common to teach in calculus classes that 0^0 is one of the "indeterminate forms", along with 0/0 and so forth, which, so the story goes, is a different thing from being undefined, like 1/0. Since, after all, if f(x) approaches 1 and g(x) approaches 0, then the limit of f(x)/g(x) is undefined, whereas if f(x) and g(x) both approach 0, then the limit of f(x)/g(x) cannot be predicted in advance. So 1/0 is undefined, but 0/0 is indeterminate.
Now this certainly is getting at a real distinction! But it's not a distinction between the value of 1/0 and that of 0/0; both are undefined. "Indeterminate" is not some separate actual value. Rather, they are getting at the distinction of the behavior of the division function near the point (1,0) vs. how it behaves near the point (0,0). Not at the points! At both those points, the function is not defined.
And similarly with 0^0. Of course, it's pretty common to define that 0^0=1, and it's a definition I'd agree with -- but this is not inconsistent with the calculus teacher's statement that 0^0 is "indeterminate", because the latter (once made sense of) is not really a statement about the value of 0^0 at all; it's a statement about how the exponentiation function behaves near the point (0,0) (not at it; at it, it's equal to 1, or at least by my definition it is, at any rate).
In short, there's no such value as "indeterminate"; the calculus teacher's "indeterminate forms" (as opposed to "undefined"), while getting at a real destinction, is not actually about the value of the function at the point at all.