Consider the following family of functions
$$f\_n(w):=\sum\_{k=0}^{\infty}\frac{(-1)^{k-1}}{k!}(k+n)^{k-1}w^k.$$



> 
> **QUESTION 1.** Does the following hold?
>  $$f\_n(w)=-\frac1{n(f\_{-1}(w))^n}.$$
> 
> 
> 


Deeper look:



> 
> **QUESTION 2.** Is there a conceptual reason why the linear shift, in $n$ of $f\_n$, translates into reciprocal power?
> 
> 
> 




---


**UPDATE.** Thanks to Stanley's answer below, I explored the web and have found very useful resources on "polynomials of binomial type" which I post for the interested reader.


[A paper by Gian-Carlo Rota et al](https://www.math.ucdavis.edu/~deloera/MISC/BIBLIOTECA/trunk/Rota1.pdf)


[Slides by Richard P. Stanley](http://www-math.mit.edu/~rstan/transparencies/bintype2)



sequencial shift on families =flipped powers. How?
It is well-known that the Abel polynomials $p\_k(x)=x(x-ak)^{k-1}$ are
a sequence of polynomials of binomial type, i.e.,
 $$ \sum\_{k\geq 0} p\_k(x)\frac{w^k}{k!} = \left( \sum\_{k\geq 0}
 p\_k(1)\frac{w^k}{k!}\right)^x, $$
which explains your formula.


