 Solution

Let $u=e^{-x^2}$, so $u^{\prime }+2xu=0$. Differentiate this identity $n$ times, for $n\ge 1$. By the Leibniz rule, only the first two derivatives of the linear factor $x$ can contribute:Rearranging provesThe case $n=0$ is the starting identity $u^{\prime }+2xu=0$, treated separately without introducing a negative derivative order.