At the CMFT international conference in Turkey (2009), the following open problem was given:
Show that $$p_n(x):=\sum_{k=0}^n \frac{(n-k)^k}{k!}x^{n-k}$$ has only real simple zeros for every $n$. Assume $p_0(x)=1$.
I have been struggling with this problem for days and as always my lack of knowledge of differential equations seems to get in the way. Letting $f^*(x)=x^{\deg(f)}\cdot f(\frac{1}{x})$, we derive the following $$\bigg(x\frac{d}{dx}p^*_{n+1}(x)\bigg)^*=x\frac{d}{dx}p_n(x) .$$ Following through the calculations we arrive at $$(n+1)p_{n+1}=xp'_{n+1}(x)+xp'_n(x).$$ Letting $F(x,t):=\sum_{k=0}^\infty p_k(x)t^k$, we arrive at the following differential equation $$t F_t = xF_x+xtF_x.$$ Please help me solve for $F$ to get the a generating function for $\{p_n\}$, that will hopefully help to solve the problem.
I think I have found the generating function. After using the formula in here provided by Raymond Manzoni and then guessing for hours. I arrived at $$\frac{1}{1-xte^t}=\sum_{k=0}^\infty p_k t^k,$$ which satisfies the differential equation and has the appropriate initial conditions. Still haven't been able to prove these have only real zeros.