limit of the sequence $x_{n}:= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n\ldots}}}$ I was thinking what happens with the sequence $\{x_n\}_{n\in \Bbb N}$ where:
$$x_{n}:= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n\ldots}}}$$
When you look some terms, for example $x_{1}=1$, $x_{2}=\sqrt[]{2 \sqrt[]{2 \sqrt[]{2 ...}}}$, $x_{3}=\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}$, these terms and the others will be continued fractions, where each one converges.
I'm asking what happens with $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}$ ?. I have an idea and it is $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}=1$. My reasoning is in the fact:
$$\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}= \displaystyle {n^{\frac{1}{n}}} n^{\frac{1}{n^2}} n^{\frac{1}{n^3}}...$$
And you know that:
$${\displaystyle \frac{1}{n}> \frac{1}{n^k} \textrm{ for } n,k \in \Bbb N}$$
Then:
$$n^{\frac{1}{n}}> n^{\frac{1}{n^k}} \geq 1$$
Like $\lim\limits_{n \to \infty}n^{\frac{1}{n}}=1$  and $\lim\limits_{n \to \infty}1=1$, by the Squeeze Theorem $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}=1$. Is this reasoning correct? What do you think about $x_{n}$? Do you think there is another way to prove it? I receive suggestions or comments. Thank you.

An Inductive idea is:
$$x_{2}=\sqrt[]{2 \sqrt[]{2 \sqrt[]{2 ...}}}\to 2\\
x_{3}=\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}\to \sqrt[2]3\\
x_{4}=\sqrt[4]{4 \sqrt[4]{4 \sqrt[4]{4 ...}}}\to \sqrt[3]4\\
x_{5}=\sqrt[5]{5 \sqrt[5]{5 \sqrt[5]{5 ...}}}\to \sqrt[4]5\\\vdots\\
x_{n}= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}\to \sqrt[n-1]n$$and it tends to $$\sqrt[n-1]n=n^{\frac{1}{n-1}}\to 1$$
Implicit : idea to solve for example$$\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}=a\to \text{to the power of 3}\\a^3=3\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}\\a^3=3\underbrace{\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}}_{a}\\a^3=3a\underbrace{\to}_{a\neq 0}a^2=3\to a=\sqrt 3$$
