Why does $\sqrt{n\sqrt{n\sqrt{n \ldots}}} = n$?

$$\begin{align} X & = \sqrt{n \cdot \sqrt {n \cdot \sqrt{n \dots} } } \\ & = \sqrt{n} \cdot \sqrt{\sqrt{n}} \cdot \sqrt{\sqrt{\sqrt{n}}} \cdot \dots \\ &= n^{1/2} \cdot n^{1/4} \cdot n^{1/8} \dots \\ &= n^{1/2 + 1/4 + 1/8 \dots} \\ &= n^1 \\ &= n \\ \end{align}$$

Suppose $y = \sqrt{x\sqrt{x\sqrt{x\cdots}}}$.

Multiply both sides by $x$ and take the square root:

$$\sqrt{xy} = \sqrt{x\sqrt{x\sqrt{x\cdots}}} = y$$

Therefore, $\sqrt{xy} = y$, and solving we have $xy = y^2 \implies x = y$.

It is important to show that the limit exists. Let define the sequence $$ a_k=\sqrt{\vphantom{A}na_{k-1}} $$ Since $\dfrac{a_k}{a_{k-1}}=\sqrt{\dfrac{n}{a_{k-1}}}$ and $\dfrac{a_k}{n}=\sqrt{\dfrac{a_{k-1}}{n}}$, 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 $$ \lim_{k\to\infty}a_k=\lim_{k\to\infty}\sqrt{\vphantom{A}na_{k-1}}=\sqrt{n\lim_{k\to\infty}a_k} $$ Squaring and dividing by $\lim_{k\to\infty}a_k$, we get that $$ \lim_{k\to\infty}a_k=n $$

Another Approach

Not as rigorous, but perhaps more intuitive. Take the logarithm of both sides and we get $$ \begin{align} \log\left(\sqrt{n\sqrt{n\sqrt{n\dots}}}\right) &=\frac12\left(\log(n)+\frac12\left(\log(n)+\frac12\left(\log(n)+\vphantom{\frac12}\dots\right)\right)\right)\\ &=\frac12\log(n)+\frac14\log(n)+\frac18\log(n)+\dots\\ &=\log(n)\left(\frac12+\frac14+\frac18+\dots\right)\\[6pt] &=\log(n) \end{align} $$

You're basically doing this:

$x_0 = \sqrt x$

$x_1 = \sqrt{x x_0}$

$\displaystyle x_n = \sqrt{x x_{n-1}} = x^{\sum_{k=1}^n \frac{1}{2^k}} $

So it's pretty obvious that converges to $x$