TITLE: "Binomial theorem"-like identities
QUESTION [30 upvotes]: There are several identities which resemble the binomial theorem.  For starters, we have the binomial theorem itself: $$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$$
But I just learned from the book "Concrete Mathematics", Exercise 5.37, that the "falling factorial" $x^{\underline{k}} = x(x-1)\ldots(x-k+1)$ satisfies a similar identity: $$(x+y)^\underline{n} = \sum_{k=0}^n \binom{n}{k} x^\underline{k} y^\underline{n-k}$$ The "rising factorial" $x^{\overline{k}} = x(x+1)\ldots(x+k-1)$ also satisfies such an  identity.
Sometimes, the identity involves a product instead of a sum on the left side.  If $f$ and $g$ are $n$-times differentiable functions on $\mathbb{R}$, then this generalization of the product rule holds: $$(fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(k)} g^{(n-k)}$$ where $f^{(k)}$ denotes the $k$-th derivative of $f$, and the 0th derivative of a function is the function itself.
Question: Are there any more of these binomial-theorem like identities in other contexts?  Are these identities part of some more general result, where we can axiomatize some conditions under which some "iterative" process satisfies a binomial-like theorem?

REPLY [8 votes]: I up-voted Qiaochu Yuan's answer.  To be more explicit: Say you have a sequence of scalars $c_1,c_2,c_3,\ldots$ (starting with $1$, not with $0$).  Then you can find a sequence of polynomials $p_0(x),p_1(x),p_2(x),\ldots$ (starting with $0$, not with $1$) such that for $n=0,1,2,3,\ldots$
$$
p_n(x+y) = \sum_{k=0}^n \binom n k p_k(x) p_{n-k}(y)
$$
and $p_n\,'(0)=c_n$ for $n\ge 1$.  This is a "polynomial sequence of binomial type".
And what I just wrote amounts to a definition by recursion, so if you want to write out, for example $p_6(x)$ with all its coefficients as polynomials in $c_1,\ldots, c_6$, just apply what I wrote above.  The polynomials in $c_1,c_2,c_3,\ldots$ that are the coefficients are the incomplete Bell polynomials, named after Eric Temple Bell.