Consider the Gaussian function $f(z)=e^{-z^2}$ which has no zeros on the complex domain. Let $D$ denote derivative w.r.t. the variable $z$.

Question.Is it true that $D^nf(z)=0$ has only real roots that are simple? If so, any slick proof?

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Consider the Gaussian function $f(z)=e^{-z^2}$ which has no zeros on the complex domain. Let $D$ denote derivative w.r.t. the variable $z$.

Question.Is it true that $D^nf(z)=0$ has only real roots that are simple? If so, any slick proof?

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The (physicists') Hermite polynomials are

$$ H_n(x) = (-1)^n e^{x^2} D^n e^{-x^2}$$

And their roots are real. For that you don't need to know they are Hermite polynomials: just Rolle's theorem. See this.

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I've found a slightly different argument.

If $f_m(z)=\left(1-\frac{z^2}m\right)^m$ then $\lim_{m\rightarrow\infty}f_m(z)=e^{-z^2}$ uniformly on every compact subset of $\mathbb{C}$. Hence, the same holds for $\lim_{m\rightarrow\infty}D^nf_m(z)=D^ne^{-z^2}$. On the other hand, $D^nf_m(z)$ has only real zeros. Therefore $D^ne^{-z^2}$ can not have non-real zeros, by Hurwitz's Theorem.

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There are several complete characterizations of real entire functions whose all derivatives have all roots real: a) this is a closure of polynomials with the same property, and b) this class is represented by the formula $$f(z)=cz^me^{-az^2+bz}\prod_{k}\left(1-\frac{z}{z_k}\right)e^{z/z_k},$$ where $a\geq0$, $b,c$ are real, $m\geq 0$ is an integer and $z_k$ real, with $$\sum\frac{1}{|z_k|^2}<\infty.$$ This is a parametric description: each such function is represented by this formula, and each function represented by this formula has the stated property. This class of function has a standard name: Polya-Wiman class.

These results are due to Wiman, Polya, Hellerstein and Williamson. For recent generalizations and survey, see arXiv:math/0510502.

manythanks @GerryMyerson -- I did not know that trick $\endgroup$ – Kevin Buzzard Jan 19 '17 at 22:54