I have been pondering the question, what does $f(x)$ look like if $f(x)=\left(\frac{d^{w}}{dx^{w}}f(x)\right)^n$ and $f(x)$ is a polynomial. If you're given w, then I believe that there are a limited number of n values and families of polynomials that satisfy this property.
I went through and solved for $w=1$:
$$ n=2; f(x)=\frac{1}{4}x^2+bx+b^2=\frac{1}{4}(x+2b)^2 $$
and $w=2$:
$$ n=3; f(x)=\frac{1}{\sqrt{216}}x^3+bx^2+2\sqrt{6}b^2x+8b^3=\frac{1}{\sqrt{216}}(x+2\sqrt{6}b)^3 $$
$$ n=2; f(x)=\frac{1}{144}x^4+bx^3+54b^2x^2+1296b^3x+11664b^4=\frac{1}{144}(x+36b)^4 $$
(b can be any real number). I see a pattern there (with the equivalent polynomial being shifted to the left or right), but I want to generalize. I'll be using $f(x)=a_{m}x^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0$ from here on out.
$$f(x)\text{ can be represented as }\sum_{k=0}^{m}a_kx^k;\text{ and }\frac{d^w}{dx^w}f(x)=\sum_{k=w}^{m}a_k\frac{k!}{(k-w)!}x^{k-w}$$
Using the Multinomial Theorem, we can write $\left(\frac{d^w}{dx^w}f(x)\right)^n$ as
$$\sum_{i_{w}+i_{w+1}+i_{w+2}+\cdots+i_{m}=n}\frac{n!}{i_{w}!i_{w+1}!i_{w+2}!\cdots i_{m}!}\left(\prod_{k=w}^{m}\left(a_k\frac{k!}{(k-w)!}x^{k-w}\right)^{i_{k}}\right)$$
I've evaluated for $a_0$, $a_1$, $a_2$, and $a_m$, but finding other terms looks like an impossible mess. $a_2$ was already difficult enough.
I want an equivalent Multinomial Theorem that allows for me to find the coefficient for any term with a certain degree without considering every way to make that degree. Hopefully, I can more easily evaluate for w or $m-1$ and find a pattern or relations between each term.