Looking for help with a proof that n-th derivative of $e^\frac{-1}{x^2} = 0$ for $x=0$. Given the function 
$$
f(x) = \left\{\begin{array}{cc}
e^{- \frac{1}{x^2}} & x \neq 0
\\
0 & x = 0
\end{array}\right.
$$
show that $\forall_{n\in \Bbb N} f^{(n)}(0) = 0$.
So I have to show that nth derivative is always equal to zero $0$. Now I guess that it is about finding some dependencies between the previous and next differential but I have yet to notice one. Could you be so kind to help me with that?
Thanks in advance!

What about a direct approach?:
$$f'(0):=\lim_{x\to 0}\frac{e^{-\frac{1}{x^2}}}{x}=\lim_{x\to 0}\frac{\frac{1}{x}}{e^{\frac{1}{x^2}}}\stackrel{\text{l'Hosp.}}=0$$
$$f''(0):=\lim_{x\to 0}\frac{\frac{2}{x^3}e^{-\frac{1}{x^2}}}{x}=\lim_{x\to 0}\frac{\frac{2}{x^4}}{e^\frac{1}{x^2}}\stackrel{\text{l'Hosp.}\times 2}=0$$
................................ Induction.................
