PI π

Note that:

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

Taking the square root of any number $n$ where $n \gt 0$ enough times will eventually converge towards $1$ .

All numbers ( Including complex numbers ) take square root many times, it must be $\boxed { 1 }$ .

And like $\sqrt { 0 \sqrt { 0 \sqrt { 0 \cdots } } } \cdots$ , $0$ does not fit the question, because taking infinite square root to $0$ , it is always $\boxed { 0 }$ .

We can write $n$ as a limit: $n=\displaystyle\lim_{x \rightarrow \infty} \pi^{.5^{x}} = \pi^0 = \boxed{1}$