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!
 A: Suppose that $R(x)$ is a rational function of $x$. Then
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\left(R(x)\,e^{-1/x^2}\right)
&=\left(R'(x)+\frac{2R(x)}{x^3}\right)e^{-1/x^2}
\end{align}
$$
where $R'(x)+\frac{2R(x)}{x^3}$ is another rational function of $x$.
Inductively, we have that there is some rational $R(x)$,
$$
\frac{\mathrm{d}^k}{\mathrm{d}x^k}e^{-1/x^2}=R(x)\,e^{-1/x^2}\tag{1}
$$

Since $R(1/x)$ is also a rational function of $x$, there is an $n$ so that
$$
\lim_{x\to\infty}\frac{R(1/x)}{x^{2n}}=0
$$
Furthermore,
$$
\begin{align}
\lim_{x\to+\infty}x^ne^{-x}
&=\lim_{x\to+\infty}\frac{x^n}{e^x}\\
&=\lim_{x\to+\infty}\frac{n!}{e^x}&&\text{applying L'Hospital $n$ times}\\[5pt]
&=0
\end{align}
$$
Now we need to show that
$$
\begin{align}
\lim_{x\to0}R(x)\,e^{-1/x^2}
&=\lim_{u\to\infty}R(1/u)\,e^{-u^2}\\
&=\lim_{u\to\infty}\frac{R(1/u)}{u^{2n}}\,\lim_{u\to\infty}u^{2n}e^{-u^2}\\
&=\lim_{u\to\infty}\frac{R(1/u)}{u^{2n}}\,\lim_{v\to+\infty}v^ne^{-v}\\[7pt]
&=0\cdot0\tag{2}
\end{align}
$$

Combining $(1)$ and $(2)$ yields
$$
\lim_{x\to0}\frac{\mathrm{d}^k}{\mathrm{d}x^k}e^{-1/x^2}=0\tag{3}
$$
A: 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.................
