$\lim\limits_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdot\cdot \cdot \cdot \{n\sqrt{2}\} } $

Since $\{ n\sqrt{2} \}$ is equidistributed modulo $2$, the limit could be rewritten as the limit of the expected value of the geometric average of $n$ uniform random variables. The integral for this would be $$\lim_{n \to \infty}\int_0^1 \int_0^1... \int_0^1 (x_1x_2...x_n)^{\frac{1}{n}} dx_n...dx_2dx_1$$

This can actually be rewritten as $$\lim_{n \to \infty}\left(\int_0^1 x^{\frac{1}{n}} dx \right)^n$$

since each $x_i$ is independent of the others. The inner integral is then equal to $\frac{n}{n+1} = 1-\frac{1}{n+1}$, so the limit is $$\lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)^{n}$$

which is clearly $e^{-1}$.