Mistake : Learning(Inspiration)

Let's assume that : $\sqrt{\sqrt[2]{\sqrt[3]{...\sqrt[n-2]{\sqrt[n-1]{n}}}}} \ge 3$ $\begin{aligned} & \implies n^{\large{\frac{1}{1\times 2\times 3\times \cdots\times (n-1)} }} \ \ ≥ 3 \\ & \implies n^{\large{\frac{1}{(n-1)!}}} \ \ ≥3 \\& \implies n \ \ ≥ 3^{(n-1)!}\\ & \implies n^{n} \ \ ≥3^{n!} \end{aligned}$ it's clear that there exists no positive integers $> 0$ for which $n^n ≥ 3^{n!}$ will be true ( can proved by induction) hence it is in contradiction to our assumption.

So , the answer is $\text{No}$ .