Since $e=\sum_{n=0}^\infty\frac{1}{n!}=1+1+\frac12(1+\frac13(1+\frac14(1+\dots)))$, we have
$$4^{e-2}=\sqrt{4\cdot\sqrt[3]{4\cdot\sqrt[4]{4\cdots}}}$$
Is there however a nice way to express the radical in the title too?
Since $e=\sum_{n=0}^\infty\frac{1}{n!}=1+1+\frac12(1+\frac13(1+\frac14(1+\dots)))$, we have
$$4^{e-2}=\sqrt{4\cdot\sqrt[3]{4\cdot\sqrt[4]{4\cdots}}}$$
Is there however a nice way to express the radical in the title too?