Find the value of $(-2)^{-(2)^{(-2)}}$. Is it 16/8/-8/none?
My attempt: $a^{-x}=\frac1{a^x}$, so, $(-2)^{-(2)^{(-2)}}=(-2)^{\frac{-1}{2^2}}=\frac{1}{(-2)^{\frac14}}$. That is, I would pick 'none of the given options' as my answer.
But it is given that answer is 16. And the comments below also suggest that the correct answer is indeed 16. So I will appreciate if one can tell the mistake in my solution.
EDIT: I've cross-checked. The statement that I have posed is same as was asked in the exam. No parsing error by me. Maybe by the question-setter, don't know.
It appears that you may have confused/misread the problem statement:
$$(-2)^{-(2)^{(-2)}} = \frac{1}{\sqrt[4]{(-2)}}\quad \neq \quad\left((-2)^{-2}\right)^{-2} = (-2)^{-2\cdot -2} = (-2)^4 = 16$$
That is, we need to be careful to note where the negation of an exponent occurs $(-2)^{a} \neq -(2^{a})$, and we need to be careful to disambiguate $(a^b)^c$ from $a^{(b^c)}.$
EDIT:
If the problem you posted is precisely what appears on the practice exam, then you've correctly argued that none of the given answers are correct. In that case, indeed, there must have been a parsing error in the source, if indeed the answer were to be $16$.