TITLE: $p_n(x)=p_{n-1}(x)+p_{n-1}^{\prime}(x)$, then all the roots of $p_k(x)$ are real
QUESTION [32 upvotes]: $p_0(x)=a_mx^m+a_{m-1}x^{m-1}+\dotsb+a_1x+a_0(a_m,\dotsc,a_1,a_0\in\Bbb R)$ is a polynomial,  and 
$$p_n(x)=p_{n-1}(x)+p_{n-1}^{\prime}(x),\qquad n=1,2,\dotsc$$
then, there exist $N\in\Bbb N$, such that $\forall n \geq N$, all the roots of $p_n(x)$ are real

I think, let $f_n(x)=e^xp_n(x)$, then 
$$f_n(x)=\frac {d(f_{n-1}(x))}{dx} $$
we get that
$$f_n(x)=\frac{d^n\left( e^xp_0(x)\right)}{dx^n}$$
so, If there exists a $p_N(x)$ such that $p_N(x)$ has  $m$ real roots, then $\forall n\geq N$,  $p_n(x)$ has  $m$ real roots.
I'd be grateful for any help

REPLY [9 votes]: Another approach.
I will prove the result step by step.


$$\sum_{j=0}^m (-1)^j \binom{m}{j}j(j-1)\cdots(j-k+1)=0$$ if $0\leqslant k\leqslant m-1$, and $m!$ if $k=m$.


Proof.
Use $j\binom{m}{j}=m\binom{m-1}{j-1}$.

2-$$\sum_{j=0}^m (-1)^j \binom{m}{j}j^k=0$$ if $0\leqslant k\leqslant m-1$, and $m!$ if $k=m$.

Proof.
Follows from Step $1$ by linearity.

3-$$\sigma_k(1,2,\ldots,j-1)$$ is polynomial of degree $2k$ on $j$ with leading coefficient $\dfrac{1}{2^k k!}$.

Proof.
This is shown by induction using Newton's identities

4-$$\sum_{j=0}^m (-1)^j\binom{m}{j}\sigma_k(1,2,\ldots,j-1)=0$$ if $2k<m$,
  and $1\times 3\times\cdots (m-1)$ if $2k=m$.

Follows from Step $2$ and $3$

5 $$\sum_{j=0}^m (-1)^j \binom{m}{j}(1-\frac{1}{n})\cdots (1-\frac{j-1}{n})$$ is equivalent 
  when $n\rightarrow +\infty $ to $(-n)^{-m/2}\times 1\times 3\times\cdots (m-1)$ if $m$ is even and, et is $o(n^{-m/2})$ if $m$ is odd.

Proof.
Just develop the product and use step $4$.

6- For $p_0(x)=x^m$, $p_n(-n)$ is equivalent to $(-n)^{m/2}\times 1 \times
 3\times\cdots\times (m-1)$ if $m$ is even, and $p_n(-n)=o(n^{m/2})$ if $m$ is odd.

Proof.
Use $p_n(x)=\sum_{j=0}^m \binom{n}{j}m(m-1)\cdots (m-j+1)x^{m-j}=\binom{m}{j}n(n-1)\cdots (n-j+1)x^{m-j}$  and substitute $x=-n$ to return to step $5$.

7-Denote $q_{m,n}(x)=n^{-m/2}p_{m,n}(-n+x\sqrt{n})$ with $p_0=x^m$
  We show by induction on $n$ as $q_ {m, n} $ converges to the Hermite polynomial $H_m $ when $n$ tends to infinity.

For $n=0$ is obvious.
Suppose the assertion true at rank <$ n $. Then she is true for $\frac{1}{m}p'_{m,0}(x)=x^{m-1}=p_{m-1,0}$. 
As the derivation commutes with the transformation $p_0\mapsto p_n$, we have $p'_{m,n}=mp_{m-1,n}$. 
It comes $q'_{m,n}=mq_{m-1,n}$, thus $q'_{m,n}$ converges to $mH_{m-1}=H'_m$ as $m$ tends to $+\infty$. 
Using step $6$ we have $q_{m,n}(0)$ converges to $H_m(0)$.

8- The result of step 7 is true for any monic polynomial of degree  $m$ instead of $p_0$.

Finally, We boils down to $p_0$ unitary. By continuity of the roots in the coefficients of a polynomial and the fact that the Hermite polynomials are divided to simple roots over $\mathbb{R}$, $p_n$ is also for $n$ large enough after step $8$.