1. We are asked to find the value of $x$ given that the maximum value of $\prod_{i=1}^n \cos \alpha_i$ under the conditions $0 \leq \alpha_i \leq \frac{\pi}{2}$ and $\prod_{i=1}^n \cot \alpha_i = 1$ is $\sqrt{x^{-n}}$. 2. Rewrite the constraint $\prod_{i=1}^n \cot \alpha_i = 1$. Since $\cot \alpha_i = \frac{\cos \alpha_i}{\sin \alpha_i}$, we have $$\prod_{i=1}^n \frac{\cos \alpha_i}{\sin \alpha_i} = 1 \implies \prod_{i=1}^n \cos \alpha_i = \prod_{i=1}^n \sin \alpha_i.$$ 3. Let $P = \prod_{i=1}^n \cos \alpha_i = \prod_{i=1}^n \sin \alpha_i$. Our goal is to maximize $P$. 4. To maximize $P$ under the symmetry of the problem and the given constraint, consider all angles equal: $\alpha_1 = \alpha_2 = \cdots = \alpha_n = \alpha$. 5. Then the constraint becomes: $$ (\cot \alpha)^n = 1 \implies \cot \alpha = 1 \implies \alpha = \frac{\pi}{4}.$$ 6. Compute $P$ for $\alpha = \frac{\pi}{4}$: $$ P = (\cos \frac{\pi}{4})^n = \left(\frac{\sqrt{2}}{2}\right)^n = (2^{-\frac{1}{2}})^n = 2^{-\frac{n}{2}}.$$ 7. The maximum value of $\prod \cos \alpha_i$ is given as $\sqrt{x^{-n}} = x^{-\frac{n}{2}}$. 8. Equate the expressions: $$ 2^{-\frac{n}{2}} = x^{-\frac{n}{2}} \implies 2 = x.$$ 9. Therefore, the value of $x$ is 2.