Showing $e^{-x^2} \in \mathcal{S}(\mathbb{R})$

I am currently studying Fourier Analysis on my own and have just been started to look at the Schwartz Space of rapidly decaying functions. One example of such functions is given in the notes that I use by $$f(x) = e^{-x^2}$$

I tried to come up with an argument that shows this is actually true. Here is what I did:

for any $m$ we have $f^{(m)}(x) = p(x)e^{-x^2}$ where $p(x) = ax^2 + bx + c$ is a second order polynomial. Hence the function $$h(x) = x^kf^{(m)} = x^kp(x)e^{-x^2}$$ is smooth. Furthermore, by repeated use of L'Hopitals Rule we have $$\lim_{x \to \pm \infty} h(x) = 0$$ Hence $h$ must be bounded by some constant $c_{k,m}$, as required.

Now, my question is whether the argument is valid and whether there is a more standard way of showing this. I hope the question does not sound too dumb! I am a newbie at this and cannot fall back on feedback via homework, so was hoping to get some comment here on my attempt. Thanks!

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You have the idea, but the problem is that the polynomial $p$ depends on $m$ (and also its degree). We can define $p_0(x)=1$ and $$\frac d{dx}(p_m(x)e^{-x^2})=p_m'(x)e^{-x^2}-2xp_m(x)e^{-x^2}=e^{-x^2}(p_m'(x)-2xp_m(x)),$$ so we have to define $p_{m+1}(x)=p_m'(x)-2xp_m(x)$. We have $\deg p_{m+1}=1+\deg p_m$ so $\deg p_m=m$ and for $k,m$ integers: $$|x^kf^{(m)}(x)|=|x^kp_m(x)e^{-x^2}|\leq \frac{|x^kp_m(x)|}{\sum_{l=0}^{m+k+1}\frac{x^l}{l!}},$$ which is bounded since $\deg (x^kp_m)=k+m<m+k+1$.
ah so I didn't analyse the behaviour of the derivatives of $e^{-x^2}$ in enough detail, thanks alot for the answser that helped a great deal! – harlekin Jan 30 '12 at 21:48
In fact the derivatives of $e^{-x^2}$ are linked to Hermite polynomials. – Davide Giraudo Jan 30 '12 at 21:50