Why does nnn…=n

${\displaystyle X=\sqrt{n\cdot \sqrt{n\cdot \sqrt{n\dots }}}}$
${\displaystyle =\sqrt{n}\cdot \sqrt{\sqrt{n}}\cdot \sqrt{\sqrt{\sqrt{n}}}\cdot \dots }$
${\displaystyle =n^{\frac{1}{2}}\cdot n^{\frac{1}{4}}\cdot n^{\frac{1}{8}}\dots }$
${\displaystyle =n^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\dots }}$
${\displaystyle =n^1}$
${\displaystyle =n}$

Suppose ${\displaystyle y=\sqrt{x\sqrt{x\sqrt{x}}}}$
Multiply both sides by x and take the square root:
$\sqrt{xy}=\sqrt{x\sqrt{x\sqrt{x...}}}=y$
Therefore, ${\displaystyle \sqrt{xy}=y}$, and solving we have ${\displaystyle xy=y^2\Rightarrow x=y}$

It is important to show that the limit exists. Let define the sequence $a_k=\sqrt{n{a}_{k-1}}$ Since $\frac{a_k}{a_{k-1}}=\sqrt{\frac{n}{a_{k-1}}}$ and $\frac{a_k}{n}=\sqrt{\frac{a_{k-1}}{n}}$, we have 1. if $a_{k-1}\le n$, then $a_{k-1}\le a_k\le n$; that is, $a_k$ is increasing and bounded above by n. 2. if $a_{k-1}\ge n$, then $a_{k-1}\ge a_k\ge n$; that is, $a_k$ is decreasing and bounded below by n. In either case, $a_k$ is convergent. Using the continuity of multiplication by a constant and the continuity of square root, we get $\underset{k\to \mathrm{\infty }}{lim}{a}_k=\underset{k\to \mathrm{\infty }}{lim}\sqrt{n{a}_{k-1}}=\sqrt{n\underset{k\to \mathrm{\infty }}{lim}{a}_k}$ Squaring and dividing by $\underset{k\to \mathrm{\infty }}{lim}{a}_k$, we get that $\underset{k\to \mathrm{\infty }}{lim}{a}_k=n$