$$\text{Calculate :}\lim_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} } . $$

**Note:**

Weyl's equidistributed criterion. The following are equivalent:
$$x_n\quad\text{is equivalent modulo 1}$$

$$\forall~ \text{continuous & 1-peridic} f: \quad\frac{1}{N}\sum_{n=1}^Nf(x_n)\rightarrow\int_0^1f $$

$$\forall~ k\in \mathbb Z^*:\quad \frac{1}{N}\sum_{n=1}^Ne^{2πikx_n}\rightarrow 0 $$

**Background:**

Im trying to approach this problem by weyl's criterion, so my thoughts so far are:

\begin{align} &\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} }\\ &=\big(\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\}\big)^{1/n}\\ &=e^{\frac{1}{n}\log\big(\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\}\big) }\\ &=e^{\big(\frac{1}{n}\sum_{k=1}^n \log(\{k\sqrt{2}\})\big)}\\ \end{align}

So, since $\sqrt{2}$ is irrational then its simple to prove that the sequence $x_n=\{ n\cdot \sqrt{2}\}$ is equidistributed $\text{mod}\ 1$.
Let us define the continuous & 1-periodic function $f(x):=\log(x-[x])$

by the weyl's criterion we get:

$$\begin{align*} \frac{1}{n}\sum_{k=1}^n \log(\{k\sqrt{2}\})&\longrightarrow\int_0^1\log(\color{black}{\underbrace{\{x\}}_{=x-[x]}})dx\\[5pt] &=\int_0^1\log(x)dx\\[5pt] &=\bigg[x\log(x)\bigg]_0^1-\int_0^1dx\\[5pt] &=-1\\[5pt] \end{align*}$$

Hence

$$\lim\limits_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} }=e^{-1} $$ Is there something wrong? Also, can we find this limit with some other way?

Let me know, thank you.