Is $0^i$ indeterminate? According to Euler's identity  


  $$e^{ix}=\cos{x}+i\sin{x}$$  


Using this identity  


  $$0^i=e^{i\ln{0}}=\cos\ln0+i\sin\ln0=\cos{-\infty}+i\sin{-\infty},\because \ln0=-\infty$$    


But $\cos{-\infty}$ and $\sin{-\infty}$ are indeterminate Does it mean that $0^i$ is indeterminate.
Please Explain.
The critical point comes from the trigonometric functions. Instead, start from the equality
$$
e^{\ln x} = x
$$
which holds for every positive real $x$. Then $\lim_{x \to 0^+} x = 0$, $\lim_{x \to 0^+} = -\infty$ and $\lim_{x \to 0^+} e^{\ln x} = 0$, since $x \to e^x$ is continuous on $\mathbb{R}$ and tends to $0$ when $x$ tends to $-\infty$. Thus in this case, $e^{-\infty} = 0$ is an acceptable convention, which is actually frequently used in mathematics.

However, in your case, neither $\cos x$ nor $\sin x$ have a limit when $x$ tends to $-\infty$. I hope this explains the difference between the two cases.
As you point out, one can not define $0^i$ via the identity of Euler. The question is whether we can find any other good properties of $0^i$. For me good properties would be that the function $f:x\mapsto x^i$ is continuous. This function is basically defined for $\mathbb{C}\backslash\mathbb{R-}$, where the $log function is not defined.

Can we extend the definition of the function to $0$?
For this we see 
$$\lim_{x\in\mathbb{R},x\rightarrow 0} x^i =$$
$$=\lim_{x\in\mathbb{R},x\rightarrow 0} e^{i\cdot log(x)}$$

This limit does not converge, so we do not have any continuous extension to $0$