Although ${0^0}$ is not necessarily defined as $1$, it is often defined that way, with reasons provided as to why. Assume, then, that it is $1$. What would the value of ${0^{0^{0...}}}$ be? (Please forgive me if the answer is just, "There is no [yet?] provable answer." I know that in tetration, you evaluate downward, so ${0^{0^0}}$ would evaluate to ${0^1}$, which would then evaluate to $0$. If you had a neverending such sequence, though, would you end up with two answers, maybe? Like, if you take the infinite sequence pairwise, each pair collapses to 1, and the resulting infinite tetration of $1$ goes to $1$. But IDK because I don't know enough about how similar (similar enough!) questions are decided, so it appears to me you could have a downward evaluation of an infinite tower that kept fluctuating between $0$ and $1$ per pair, maybe even terminating at $0$. Based on a (hazy) recollection of Conway and Guy's The Book of Numbers re: $\omega$ (how to characterize odd and even numbers using $\omega$), I suppose that maybe you'd say the tower would collapse to $0$ if there were $\omega+1$ steps (the extra step rendering the final computation incapable of being done pairwise without remainder)? (I'm just shooting in the dark here.)
I tried to find more about this stuff elsewhere beforehand, but didn't come up with anything. Apologies if there's some essay or book out there that I should know about already, that goes over this question.
Whether $0^0$ is defined or not does not really matter. The only real numbers for which the infinite tetration converges have a strictly positive lower bound ($e^{-e}$) or else the specific value $-1$.