Though $0^0$ is an indeterminate form of limits, is it undefined? I’ve done a my research, though I have not been able to find an adequate explanation as to whether or not $$0^0$$ exists as a real number, and why or why not? I must credit this question to “Question on the controversial ‘undefined’ $0^0$.”
This Wikipedia entry lists a (seemingly?) exhaustive list of indeterminate forms of limits, with which I take no issue. All of which involve $\infty$ or a ‘multiple’ of $1/0$ and are therefore undefined as real numbers—that is, all except $0^0$.
Desmos clearly claims that $0^0=1$; however, my TI-84 returns ERROR: DOMAIN in both real and complex mode. So, which is it? Is $0^0$ defined or not?
 I wasn’t quite sure what other tags apply, so feel free to edit.

Zero to zeroth power is often said to be "an indeterminate form", because it could have several different values. 
Since $x^0$ is $1$ for all numbers $x$ other than $0$, it would be logical to define that $0^0 = 1$. 
But we could also think of $0^0$ having the value $0$, because zero to any power (other than the zero power) is zero. 
Also, the logarithm of $0^0$ would be $0$ · infinity, which is in itself an indeterminate form. So laws of logarithms wouldn't work with it. 
So because of these problems, zero to zeroth power is usually said to be indeterminate. 
However, if zero to zeroth power needs to be defined to have some value, $1$ is the most logical definition for its value. This can be "handy" if you need some result to work in all cases (such as the binomial theorem).
