Sum of derivatives of a polynomial

I have to admit the following solution was proposed to me by a friend, I did not find it:

Let $$f(x) = p(x) + p^\prime(x) + ... + p^{(n)}(x) $$

Note that $$ f^\prime = p^\prime + p^{\prime \prime} + ...+ p^{(n)} $$

that is, $$ f = p + f^\prime$$

Clearly $n$ is even. Hence $f$ has even degree, too. This implies that $f$ attains it's absolute minimum (it's not a maximum, as $f$ behaves like $p$ at infinity) in some point $z_0$, hence $ f^\prime(z_0) = 0.$ Consequently, $\forall x$, $$f(x) \ge \min(f) = f(z_0) = p(z_0) + f^\prime (z_0) = p(z_0) \ge 0 $$

Instead of $L_n := (1 + D + D^2 + \dots + D^{n})$, use $L_{\infty} := (1 + D + D^2 + \dots)$, which comes to the same thing for polynomials of degree $n$ or less. If we let

$$\sigma(x) = p(x)+p'(x)+p''(x)+\dots+p^{(n)}(x) = L_{\infty}(p)(x)$$

then we can sum the geometric progression in $D$ to get

$$\sigma(x) = ((1 + D + D^2 + \dots)p)(x) = ((1-D)^{-1}p)(x)$$

Thus

$$p(x) = ((1-D)\sigma)(x)$$

This might look like trickery, but if you check it against the original expression for $\sigma$ you can see that it works:

$$((1-D)\sigma)(x) = p(x)+p'(x)+p''(x)+\dots+p^{(n)}(x) - (p'(x)+p''(x)+\dots+p^{(n)}(x))$$ $$= p(x)$$

Now define $\tau(x) = e^{-x}\sigma(x)$, so $\tau'(x) = e^{-x}(\sigma'(x) - \sigma(x)) = e^{-x}((D-1)\sigma)(x) = -e^{-x}p(x)$. By hypothesis, this is $ \le 0$ for all $x \in \mathbb R$. Also, $\sigma$ is a polynomial, so $\tau(x) \to 0$ as $x \to \infty$. Therefore $\tau(x) \ge 0$ for all $x$.

Hence $\sigma(x) \ge 0$ for all $x$, which is what we wanted.