Some keywords you'll want to look into: binomial type, Appell sequence, Sheffer sequence, umbral calculus. The references in the corresponding Wikipedia articles are good too.
Edit: In some sense, all of these identities can be deduced from the last one. Setting
$$f(t) = e^{xt}, g(t) = e^{yt}$$
produces the binomial theorem, and setting
$$f(t) = (1 + t)^x = \exp (x \log (1 + t)), g(t) = (1 + t)^y = \exp ( y \log (1 + t))$$
produces the second identity. From this perspective one can think of the study of generalized binomial theorems as being all about generating functions of the form $\exp (x h(t))$ where $h(0) = 0$; setting
$$f(t) = \exp (x h(t)), g(t) = \exp (y h(t))$$
produces a fairly general binomial theorem, especially if one writes $h(t) = \sum_{n \ge 1} h_n t^n$ as a formal power series in formal variables.
I hope I understood your question correctly.
For any sequence of real numbers $(\alpha_n)_{n\geq 0}$ and $p\in\mathbb N$ let us denote by $(\alpha_n^{*p})$ (awkward notation) the sequence defined by
$$\alpha_n^{*p}=\sum_{k=0}^n \left(\frac{n^{\underline k}}{n^k}\right)^p \alpha_k\, . $$
Then, the following holds true: for any absolutely convergent series $\sum\alpha_k$, the sequence $(\alpha_n^{*p})_{n\geq 0}$ is convergent with limit $\sum_0^\infty\alpha_k$. In particular, with $\alpha_k=\frac{(-1)^k}{k!}$ and $p=2$ you get that $c_n\to1/e$.
To see this, note that one can write
$$\alpha_n^{*p}=\sum_{k=0}^\infty q_{n,k}\, \alpha_k\, ,$$
where
$$q_{n,k}=\left\{ \begin{matrix}
\left(\frac{n^{\underline k}}{n^k}\right)^p&k\leq n\\0&k>n
\end{matrix}\right. $$
For each fixed $k$ we have $\lim_{n\to\infty} q_{n,k}=1$ by the formula for $\frac{n^{\underline k}}{n^k}$ you give in your question. Moreover, since $0\leq q_{n,k}\leq 1$ by the same formula, we also have $\vert q_{n,k}\alpha_k\vert\leq \vert\alpha_k\vert$ for all $n$ and every $k\geq 0$. By the dominated convergence theorem (for series), it follows that
$$\lim_{n\to\infty} \sum_{k=0}^\infty q_{n,k}\alpha_k=\sum_{k=0}^\infty\alpha_k\, , $$
which is the required result.
Best Answer
How many ways can you pick $k$ balls from a set of $n$ different red balls and $m$ green different balls?
Answer
$$(n+m)^{\underline k}$$
But you can count them another way. First suppose that the $k$ balls are red, then $k-1$ are red and $1$ is green, etc.
This gives
$$\sum_{j=0}^k\binom kj n^{\underline j} m^{\underline {k-j}}$$