The square root of the square root of the square root...

You should know that:

$\Large{\sqrt{2015}=2015^{\frac{1}{2}}}$

Then:

$\Large{\sqrt{\sqrt{\sqrt{\sqrt{\ldots 2015}}}}}$

$\Large{=2015^{(\frac{1}{2} \times \frac{1}{2} \times\frac{1}{2} \times\frac{1}{2} \ldots)}}$

$\Large{= 2015^{\frac{1}{\infty}}=2015^0=\boxed{1}}$

We note that when $x > 1$ , we note that: $x > 1 \quad \Rightarrow \sqrt{x} > \sqrt{1} \quad \Rightarrow \sqrt{\sqrt{x}} > \sqrt{\sqrt{1}} \quad \Rightarrow ... \quad \Rightarrow \sqrt{\sqrt{\sqrt{...\sqrt {x}}}} = 1$

When $x<1$ , then $\sqrt{\sqrt{\sqrt{...\sqrt {x}}}} = 0$

It can also be shown by letting $y = \sqrt{\sqrt{\sqrt{\sqrt{...x}}}}$ , then be have:

$\begin{aligned} y & = \sqrt{\sqrt{\sqrt{\sqrt{...x}}}} \\ y^2 & = \sqrt{\sqrt{\sqrt{\sqrt{...x}}}} \\ \Rightarrow y^2 & = y \\ \Rightarrow y & = \begin{cases} 0 & \text{for } x < 0 \\ 1 & \text{for } x \ge 1 \end{cases} \end{aligned}$

This is equal to $\lim_{n\to\infty}x^{2^{-n}}$ . Since $\lim_{n\to\infty}2^{-n}=0$ , then for all $x \in R^{+}$ , the expression equals $x^0=1$ . Note that $N \subseteq R^{+}$ .

Square root reduces the value and here we have infinite square roots. The least value of a number when you apply square root is 1.

So infinite square root leads to least value that is 1

$\sqrt { \sqrt { \sqrt { \dots \sqrt { 2015 } } } } = 2015^\frac { 1 }{ \infty }$

$\lim _{ x\rightarrow \infty }{ \frac { 1 }{ \infty } } =0$

$2015^0=1$

I got the correct answer, but I'm so glad to see the actual solutions because honestly I just guessed.