sequencial shift on families =flipped powers. How? 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
Slides by Richard P. Stanley
 A: 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.
A: It is also worth observing that $W(z):=-zf_1(z)$ is the  Lambert function, the inverse function at $z=0$ of $ze^z$, whose  power series expansion can be deduced by means of the Lagrange inversion theorem; and the expansion of any  power of it as well, from which your formula follows. 
A: This is unusual but I've also found an answer inspired by the above solutions.
Abel's formula 
$$\sum_{k=0}^n\binom{n}kx(x-kz)^{k-1}(y+nz)(y+kz)^{n-k-1}=(x+y+nz)(x+y)^{n-1}.$$
Lemma. Denote $c=f_{-1}(w)$ and let $i\in\Bbb{Z}$. Except for $f_0(w)$, we have the identities 
$$\frac{if_i(w)}{(i+1)f_{i+1}(w)}=c \qquad\text{or} \qquad f_i(w)=-\frac1{ic^i}.$$
Proof. For $(i+1)cf_{i+1}=if_i$, apply Cauchy's product formula:
$$\begin{align} (i+1)\,c\, f_{i+1}&=(i+1)\sum_k\frac{(-1)^{k-1}}{k!}(k-1)^{k-1}w^k\sum_k\frac{(-1)^{k-1}}{k!}(k+i+1)^{k-1}w^k \\
&=(i+1)\sum_n\frac{(-1)^nw^n}{n!}\sum_{k=0}^n\binom{n}k(k-1)^{k-1}(n+i+1-k)^{n-k-1} \\
&=i\sum_n\frac{(-1)^{n-1}w^n}{n!}(n+i)^{n-1}=i\cdot f_i; \end{align}$$
where Abel's formula has been used with $x=-1, y=n+i+1, z=-1$. For the second part, it suffices to verify $cf_1(w)=-1$ and iterate applying the part we just proved. But, $cf_1(w)=-1$ is immediate from Abel's formula (choose $x=-1, y=n+1, z=-1$). The proof is complete. 
