Prove that $F_{2n}(x)=0$ has exactly one root in the interval $x\in(0,1),$ and this root $\to 0$ when $n \to \infty.$

Question

Define $$f_1(x)=x\\f_2(x)=x^x\\\vdots\\f_{n+1}(x)=x^{f_n(x)}$$

Let $F_n(x)=f_n^{'}(x).$ Hence $$F_1(x)=1\\F_2(x)=x^x(1+\log(x))\\\vdots$$

Prove that $F_{2n}(x)=0$ has exactly one root in the interval $x\in(0,1),$ and this root $\to 0$ when $n \to \infty.$

Here are the images of $f_{2n}(x)$ for $n=1,2,\ldots,10.$