Do these polynomials have alternating coefficients?

In answering another MathOverflow question, I stumbled across the sequence of polynomials $Q_n(p)$ defined by the recurrence $$Q_n(p) = 1-\sum_{k=2}^{n-1} \binom{n-2}{k-2}(1-p)^{k(n-k)}Q_k(p).$$ Thus:

$Q_{2}(p) = 1$

$Q_{3}(p) = -p^2 + 2 p$

$Q_{4}(p) = -2 p^5 + 9 p^4 - 14 p^3 + 8 p^2$

$Q_{5}(p) = 6 p^9 - 48 p^8 + 162 p^7 - 298 p^6 + 318 p^5 - 189 p^4 + 50 p^3$

Numerical calculations up to $n=60$ suggest that:

1. The lowest-degree term of $Q_n(p)$ is $2n^{n-3}p^{n-2}$.
2. The coefficients of $Q_n(p)$ alternate in sign.

Are these true for all $n$?

As the title indicates, I'm especially puzzled about 2. Indeed, the original inspiration for the polynomials $Q_n(p)$ comes from a classic paper of E. N. Gilbert (Random graphs, Ann. Math. Stat. 30, 1141-1144 (1959); ZBL0168.40801) where the author studies the sequence of polynomials $P_n(p)$ given by the similar recurrence $$P_n(p) = 1 - \sum_{k=1}^{n-1} \binom{n-1}{k-1}(1-p)^{k(n-k)}P_k(p),$$ which do not have alternating coefficients.

• Don't know how this might help but in fact it seems that $Q_n(1-x)$ is $(1-x)^{n-2}(1+x)$ times a polynomial with positive integer coefficients and constant term 1 (with leading coefficient $(n-2)!$ and of degree $\frac{n(n-3)}2$ starting from $n=4$) – მამუკა ჯიბლაძე Oct 8 '17 at 19:01
• Experimentally, are the absolute values moreover unimodal? – Wolfgang Oct 8 '17 at 20:23
• Perhaps the stronger result is true that every zero of $Q_n(x)$ has nonnegative real part. I have only checked up to $n=10$. – Richard Stanley Oct 9 '17 at 0:41
• R. Stanley's suggestion with the observations below suggest that your polynomial (after a rotation) has all roots in the lower half plane. Such polynomials are called 'stable', and there is a bunch of results on how to prove stability, see e.g. works of P. Bränden and J. Borcea. – Per Alexandersson Oct 9 '17 at 9:58

To illustrate the suggestion of Richard Stanley about positivity of real parts of zeroes, here are the zeroes of $Q_{20}$. The pattern seems to be the same for all of them.
Another empirical observation: seems that $$\frac{Q_n(1-x)}{(1-x)^{n-2}(1+x)}=1+(n-3)x+\left(\binom{n-2}2+1\right)x^2 +\left(\binom{n-1}3+n-3\right)x^3+\left(\binom n4+\binom{n-2}2+1\right)x^4+...+\left(\binom{n+k-4}k+\binom{n+k-6}{k-2}+\binom{n+k-8}{k-4}+...\right)x^k+O(x^{k+1})$$for $n>k+1$