= Solution
Let $u=e^{-x^2}$, so $u'+2xu=0$. Differentiate this identity $n$ times, for $n\geq1$. By the <Leibniz rule>, only the first two <derivatives> of the linear factor $x$ can contribute:
$$
u^{(n+1)}+2xu^{(n)}+2nu^{(n-1)}=0.
$$
Rearranging proves
$$
\boxed{\left(2x+\frac d{dx}\right)\frac{d^n}{dx^n}e^{-x^2}
=-2n\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}}.
$$
The case $n=0$ is the starting identity $u'+2xu=0$, treated separately without introducing a negative <derivative> order.
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