Power rooting

$\begin{aligned} \Large {\left({\left(\sqrt[x^x]{4^4}\right)}^{4x}\right)}^{4x} & = \Large 4^x\\ \\ \Large 4^{4 × \tfrac{1}{x^x} × 4x × 4x} & = \Large 4^x\\ \\ \Large 4^{\tfrac{64x^2}{x^x}} & = \Large 4^x\\ \\ \Large 4^{64x^{2 - x}} & = \Large 4^x\\ \\ \text{Equating the powers}:-\\ \large 64x^{2 - x} & = \large x\\ \\ \large 64 & = \Large \dfrac{x}{x^{2 - x}}\\ \\ \Large x^{1 - (2 - x)} & = \large 64\\ \\ \Large x^{1 - 2 + x} & = \large 64\\ \\ \Large x^{x - 1} & = \large \boxed{64} \end{aligned}$