Solution:
We claim we can find values $a$ and $b$ such that $p(x) = (x - a)(x + b)^n$ is a polynomial of degree $n+1$ that satisfies these constraints. We show that its coefficients are decreasing by finding a general formula for the coefficient of $x^k$.

The coefficient of $x^k$ is $b^k \binom{n}{k} - a b^{k-1} \binom{n}{k-1}$, which can be seen by expanding out $(x + b)^n$ and then multiplying by $(x - a)$. Then we must prove that
$$b^{k+1} \binom{n}{k+1} - a b^k \binom{n}{k} < b^k \binom{n}{k} - a b^{k-1} \binom{n}{k-1}$$
or
$$a b^{k-1} \left(b \binom{n}{k} - \binom{n}{k-1}\right) > b^k \left(b \binom{n}{k+1} - \binom{n}{k}\right) .$$
Choose $b > \max \left(\frac{\binom{n}{k}}{\binom{n}{k-1}}\right)$ in order to make sure the right-hand term in each product on each side of the inequality sign is positive (we'll be dividing by it, so this makes things much easier), and choose $a > \max \left(\frac{b\left(b \binom{n}{k+1} - \binom{n}{k}\right)}{b \binom{n}{k} - \binom{n}{k-1}}\right)$ to make sure the inequality always holds. Since there are only finite values that $k$ can take on given a fixed $n$ (namely, integers between 0 and $n$ inclusive), we can always find values of $a$ and $b$ that satisfy these constraints.