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.
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.
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.
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.