In a forward difference formula, $D_-$ refers to the backward difference operator. Thus, $(D_- u)(x) = u(x)-u(x-h)$.
In the answer key of a problem that I was working on, one of the steps is:
$[D_- + \frac{1}{2}D_-^2+\frac{1}{3}D_-^3+O(h^4)]^s$
=$D_-^s + {s \choose 1}\frac{1}{2}D_-^{s+1}+ ({s \choose 1}\frac{1}{3}+{s \choose 2}\frac{1}{2^2})D_-^{s+2}+O(h^{s+3})$
Unfortunately, I don't know how they got the second expression from the first.
Aside from my own experimentation, I've checked my textbook which has a similar expression. Unfortunately, it doesn't show the derivation.
In the $s$ factors, you need to pick $D$ all $s$ times in $$\binom{s}{s}=1$$ ways to get $D^s$.
To get $D^{s+1}$, you need to pick a $D^2$ once in $$\binom{s}{1}$$ ways and the $1/2$ will appear once in the term.
To get $D^{s+2}$ you either pick two $D^2$ in $$\binom{s}{2}$$ ways with a factor of $(1/2)^2$ or you pick one $D^3$ in $$\binom{s}{1}$$ ways with one factor of $1/3$.
All other terms involve $D^{s+3}$ or greater so the equality follows.