Prove $\lim_{n \to \infty}\int_0^1 \dots \int_0^1 f(\sqrt[n]{x_1\dots x_n})dx_1\dots dx_n = f(\frac{1}{e}).$ $f$ is continuous on $[0;1].$

A short solution could be. Consider the random variables $u_1,...,u_n$ which they are i.i.d with density function $\rho(u)=e^{-u}$ (exponential distribution with $\lambda =1 $). So the expected value $E[u] = \int^{\infty}_{0}u\cdot e^{-u}du = 1$. By law of large numbers $\frac{S_n}{n}\to E[u]=1$. Where $S_n := u_1+..+u_n$. Observe that $$\int_{0}^1\dots\int_{0}^1f(\sqrt[n]{x_1...x_n})dx_1\dots dx_n = \int_{0}^{\infty}\dots\int_{0}^{\infty}f(e^{-\frac{S_n}{n}}) e^{-u_1}du_1...e^{-u_n}du_n =E[f(e^{-\frac{S_n}{n}})]$$ Finally by continuity and dominated convergence $$E[f(e^{-\frac{S_n}{n}})] \xrightarrow{n} E[f(e^{-1})]=f(e^{-1}).$$

Here's an elementary proof. Suppose $f(x) = x^k$ then we see that \begin{align} \int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n})=&\ \int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ \left(\sqrt[n]{x_1\cdots x_n} \right)^k\\ =&\ \left(\int^1_0 dx\ x^{\frac{k}{n}} \right)^n = \left(1+\frac{k}{n} \right)^{-n}. \end{align} In particular, it follows \begin{align} \lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) = \lim_{n\rightarrow \infty}\left(1+\frac{k}{n} \right)^{-n} = \left(\frac{1}{e}\right)^k. \end{align} If $f$ is a polynomial, then it follows \begin{align} \lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) = f\left( \frac{1}{e}\right). \end{align}

Next, if $f$ is continuous, then, by Wierestrass approximation theorem, there exists a sequence of polynomials $p_m$ such that $p_m \rightarrow f$ uniformly on $[0, 1]$. Finally, it follows \begin{align} \lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ f(\sqrt[n]{x_1\cdots x_n}) =&\ \lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ \lim_{m\rightarrow \infty}p_m(\sqrt[n]{x_1\cdots x_n})\\ =&\ \lim_{n\rightarrow \infty}\lim_{m\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ p_m(\sqrt[n]{x_1\cdots x_n})\\ =&\ \lim_{m\rightarrow \infty}\lim_{n\rightarrow \infty}\int^1_0\cdots \int^1_0 dx_1\cdots dx_n\ p_m(\sqrt[n]{x_1\cdots x_n})\\ =&\ \lim_{m\rightarrow \infty} p_m\left(\frac{1}{e} \right) = f\left(\frac{1}{e} \right). \end{align}