On a comment to my this answer, there is a comment where this expression is written:
$$\frac{1}{n^n} \left(p\frac{{\rm d}}{{\rm d}p}\right)^k \left(q\frac{{\rm d}}{{\rm d}q}\right)^{n-k} (p+q)^n \Bigg|_{q=1-p}$$
What does it mean? I've never seen differential operators in binomial/combinatorics context before.
By definition \begin{align} \left(q\frac{d}{dq}\right) u &= q\frac{du}{dq}, \\ \left(q\frac{d}{dq}\right)^2 u &= q\frac{d}{dq}\left(q\frac{du}{dq}\right) =q\frac{du}{dq}+q^2\frac{d^2 u}{dq^2}, \\ \left(q\frac{d}{dq}\right)^3 u &= q\frac{d}{dq}\left(\left(q\frac{d}{dq}\right)^2 u\right) = q\frac{du}{dq}+3q^2\frac{d^2u}{dq^2}+q^3\frac{d^3 u}{dq^3}, \end{align} and, recursively, $$ \left(q\frac{d}{dq}\right)^{n+1} u = \left(q\frac{d}{dq}\right)\left(q\frac{d}{dq}\right)^{n} u $$