Infinite Square Roots.

Let $x=\sqrt{3\sqrt{3\sqrt{3\sqrt{3 \cdots}}}}$ . Then

$\begin{aligned} x & = \sqrt{3x} & \small \color{#3D99F6} \text{Squaring both sides.} \\ x^2 & = 3x & \small \color{#3D99F6} \text{Since }x \ne 0 \\ \implies x & = 3 \\ \implies n & = \boxed 1 \end{aligned}$

This is quite obvious if we consider the following:

$\sqrt{3\sqrt{3\sqrt{3\sqrt{3 \cdots}}}} = \sqrt{3\sqrt{3\sqrt{3^2}}} = \sqrt{3\sqrt{3^2}} = \sqrt{3^2} = 3$

Notice that $n= \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots = \frac{a_1}{1-r}$ , where $a_1$ is the first term of an infinite geometric series and $r$ is its ratio; therefore the mentioned sum will be : $n=\sum_{n=\frac{1}{2}}^{\infty}a_n=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1.$

It is given that $\sqrt{3 \sqrt {3 \sqrt{3 \sqrt{ 3 \cdots }}}} = 3^n \: \: \:(1)$

Squaring both sides gives us $3 \sqrt {3 \sqrt{3 \sqrt{ 3 \cdots }}} = 3^{2n}$ Dividing by 3 gives us $\sqrt{3 \sqrt{ 3 \sqrt{ 3 \cdots }}} = 3^{2n-1} \: \: \: (2)$ We notice that (2) is exactly the same as (1). So it is legitimate to say $3^n = 3^{2n-1} \Leftrightarrow n = 2n-1$ Thus, we can say that $n= \boxed{1}$ .

$\sqrt{3{\sqrt{3{\sqrt{3{\sqrt{3{\dots}}}}}}}}=3^n=\sqrt{3 \times 3^n} \implies 3^{n+1}=3^{2n} \implies n+1=2n \implies n=\boxed{\large{1}}$

$\sqrt{3{\color{#3D99F6}\sqrt{3\sqrt{3\sqrt{3 ...}}}}} = \color{#3D99F6}3^n$

$\implies \sqrt{3 \times {\color{#3D99F6}3^n}} = 3^n$

$\implies 3 \times 3^n = (3^n)^2$

$3^n = 3^1$

$\boxed{n = 1}$

Let $x=\sqrt{3{\sqrt{3{\sqrt{3{\sqrt{3{\dots}}}}}}}}$

Then we can replace the part right after the first '3' with x:

$x=\sqrt{3{x}}$

And since stated in the problem that $x=3^n$

$3^n=\sqrt{3\cdot3^n}$

Simplify the right hand side:

$3^n=\sqrt{3^{n+1}}$

Since $\sqrt{n}=n^\frac{1}{2}$ ,

$3^n=3^\frac{n+1}{2}$

Get rid of the base (use $\log$

$n=\frac{n+1}{2}$

Multiply both sides by 2

$2n=n+1$

And the last step would be to subtract $n$ from both sides.

$n=1$

Therefore, $\sqrt{3{\sqrt{3{\sqrt{3{\sqrt{3{\dots}}}}}}}}=3$