Inspired by Otto Bretscher

It is given that $\large{x^{x^{\cdot^{\cdot^{x^\pi}}}}} = \pi$ . Now, if a real value $x$ satisfying the equation exists, then we have:

$\begin{aligned} x^\pi & = \pi \\ \implies x & = \sqrt[\pi]{\pi} \end{aligned}$

Now, if we consider $a_0 = x^\pi$ and $a_{n+1} = x^{a_n}$ for $n \ge 1$ , then $\displaystyle \lim_{n \to \infty} a_n = \large{x^{x^{\cdot^{\cdot^{x^\pi}}}}}$ . We note that if $x= \sqrt[\pi]{\pi}$ then $a_0 = (\sqrt[\pi]{\pi})^\pi = \pi$ and then $a_1 = a_2 = a_3 = \cdots = \pi$ , that is $\displaystyle \lim_{n \to \infty} a_n = \large{x^{x^{\cdot^ {\cdot^{x^\pi}}}}} = \pi$ . But if $x \ne \sqrt[\pi]{\pi}$ , $\displaystyle \lim_{n \to \infty} a_n$ does not converge and it has no solution. Therefore, the only solution is $x = \boxed{\sqrt[\pi]{\pi}}$ .