Gaussian Integrals Mathematica tells me:
$$\int_{-\infty}^{\infty}\frac{\mathrm{d}^n}{\mathrm{d}x^n}e^{-x^2}\mathrm{d}x\stackrel{n\neq0}{\equiv}0$$
My investigations give me on the one hand:
$$\int_{-\infty}^{\infty}\frac{\mathrm{d}^n}{\mathrm{d}x^n}e^{-x^2}\mathrm{d}x=\left|\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}e^{-x^2}\right|_{-\infty}^{\infty}=\left|p_{n-1}(x)e^{-x^2}\right|_{-\infty}^{\infty}=0$$
and on the other hand they yield:
$$\int_{-\infty}^{\infty}x^{2n}e^{-x^2}\mathrm{d}x
=\int_{-\infty}^{\infty}(-1)^n\left.\frac{\mathrm{d}^n}{\mathrm{d}a^n}\right|_{a=1}e^{-ax^2}\mathrm{d}x\\=(-1)^n\left.\frac{\mathrm{d}^n}{\mathrm{d}a^n}\right|_{a=1}\int_{-\infty}^{\infty}e^{-ax^2}\mathrm{d}x=(-1)^n\left.\frac{\mathrm{d}^n}{\mathrm{d}a^n}\right|_{a=1}\sqrt{\frac{\pi}{a}}=\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2^n}\sqrt{\pi}$$
Intuitively, the latter seems right also because I can't imagine all integrals to vanish, or do they?
(Ok my problem got solved: I had an error in computing the derivatives.)
 A: $e^{-x^2}$ belongs to the Schwarz space $\mathfrak{S}(\mathbb{R})$, hence your claim just follows from integration by parts, or from:
$$\int_{-M}^{M}\frac{d^m}{dx^m}e^{-x^2}\,dx = \left.\frac{d^{m-1}}{dx^{m-1}}e^{-x^2}\right|_{-M}^{M}\ll\frac{1}{M}.$$
A: By the main theorem of integral calculus, your integral is $\big[\frac{d^{n-1}}{d^{n-1}x}e^{-x^2}\big]_{-\infty}^{\infty}$.
But the $n-1$-derivative just gives you a linear combination of polynomials times the Gaussian function,so 
$I=\sum_{i=0}^{n-1}\left[a_ix^ie^{-x^2}\right]_{-\infty}^{\infty}$ 
Now, note that $e^{-x^2}$ decreases faster to zero then every polynomial at $\pm\infty$, which means $\lim_{x\rightarrow\infty}x^i e^{-x^2}=0$,for $i\geq0 $.
So every term of the sum vanishes independently, which means that the whole integral is zero.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{}$
\begin{align}&\color{#66f}{\large\left.%
\int_{-\infty}^{\infty}\totald[n]{\expo{-x^{2}}}{x}\,\dd x\,
\right\vert_{\,n\ \geq\ 1}}
=\lim_{y\ \to 0}\ \int_{-\infty}^{\infty}
\totald[n]{\expo{-\pars{x + y}^{\,2}}}{x}\,\dd x
=\lim_{y\ \to 0}\ \int_{-\infty}^{\infty}
\totald[n]{\expo{-\pars{x + y}^{\,2}}}{y}\,\dd x
\\[5mm]&=\lim_{y\ \to 0}\ \partiald[n]{}{y}\int_{-\infty}^{\infty}
\expo{-\pars{x + y}^{\, 2}}\,\dd x
=\lim_{y\ \to 0}\ \partiald[n]{}{y}\int_{-\infty}^{\infty}
\expo{-x^{2}}\,\dd x
=\lim_{y\ \to 0}\ 0=\color{#66f}{\large 0}
\end{align}
