Compute the limit of the sequence $x_n=(2-\sqrt{2})(2-\sqrt[3]{2})...(2-\sqrt[n]{2})$

Taking logs, any term of $$\sum _{k=2}^{\infty } \log \left(2-2^{1/k}\right)$$ is negative and decreasing.

In other words, as $$\log (2-2^x)=-x \log (2)+O\left(x^2\right);\;\text{ as } x\to 0$$ we have $$\log \left(2-2^{1/k}\right)\sim -\frac{\log (2)}{k};\;\text{ as }k\to\infty$$ The sum is equivalent to $$(-\log 2)\sum_{k=2}^{\infty}\frac1k$$ diverges to $-\infty$ thus the product converges to $0$.

We have \begin{align} \ln(x_n) &=\sum_{i=1}^n\ln(2-2^{\frac{1}{n}}) \\ \iff\frac{1}{n}\ln(x_n) &=\frac{1}{n}\sum_{i=1}^n\ln(2-2^{\frac{1}{n}}) \tag{1}\\ \end{align} And we have \begin{align} \frac{1}{n}\sum_{i=1}^n\ln(2-2^{\frac{1}{n}}) \longrightarrow \int_0^1\ln(2-2^x)dx \tag{2}\\ \end{align} Using Mathematica, I found that $$\int_0^1\ln(2-2^x)dx=\frac{\ln(2)}{2} - \frac{\pi^2}{\ln(4096)} \approx -0.839996 \tag{3}$$ From (1), (2) and (3), we have $$\ln(x_n) = n \left( \frac{\ln(2)}{2} - \frac{\pi^2}{\ln(4096)} \right) \longrightarrow -\infty $$ or $$x_n \longrightarrow 0$$