calculate limit of $a_{n+1}=e^{-a_n}$

Question

I have proven that $e^{-x}=x$ have only one solution in $(0,1)$.

How do I prove that $a_{n+1}=e^{-a_n}$, ($a_0=1$) converge to that solution?

Answer 1

hint

Let $L $ be the unique root.

then

$$|a_n-L|=|e^{-a_{n-1}}-e^{-L} |$$ $$=|a_{n-1}-L|e^{-c} $$