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?
> 
> 
> 



roots of higher derivatives of exponential
The (physicists') [Hermite polynomials](https://en.wikipedia.org/wiki/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](https://math.stackexchange.com/questions/104845/the-roots-of-hermite-polynomials-are-all-real).


