Define a function $f : \mathbb R \longrightarrow \mathbb R$ by $$f(x)=\begin{cases}e^{-1/x^2}&x\neq 0\\ 0&x=0 \end{cases}$$
Show that $f^{(n)} (0) = 0,$ for all $n \geq 1.$
I can prove that $f'(0) = f''(0) = 0.$ But for larger $n$ how do I show that? I think we have to use induction on $n.$ But I can't able to show that $f^{(n+1)} (0) = 0$ whenever $f^{(n)} (0) = 0.$ I am thinking about applying Leibnitz rule here. Is there any clever way to show that?
Some suggestion will be boon for me at this stage. Thanks!
Hint: Show that $\lim_{x\to 0}F(x)e^{-1/x^2} = 0$ for all $F\in \mathbb R(X)$ (rational fractions in $X$). Then use induction.