TITLE: A "binomial" generalization of harmonic numbers
QUESTION [7 upvotes]: For positive integers $s$ and $n$ (let's limit the generality), define
$$H_s(n)=\sum_{k=1}^{n}\frac{1}{k^s},\qquad G_s(n)=\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{k-1}}{k^s}.$$
The former is well-known; the latter is what I'm wondering about. I met it in a number of contexts. Say, if $X_1,\ldots,X_n$ are independent random variables distributed exponentially with parameter $1$, then $s!G_s(n)$ is the $s$-th moment of $\max\{X_1,\ldots,X_n\}$.
I was looking for an expression of $G_s(n)$ in terms of $H_s(n)$. Here are examples:
\begin{align}
   G_1(n) &= H_1(n),
\\ G_2(n) &= (H_2(n)+H_1^2(n))/2,
\\ G_3(n) &= (2H_3(n)+3H_2(n)H_1(n)+H_1^3(n))/6,
\\ G_4(n) &= (6H_4(n)+8H_3(n)H_1(n)+3H_2^2(n)+6H_2(n)H_1^2(n)+H_1^4(n))/24
\end{align}
(the first one is seen in the linked article and in some questions on this site).
The general formula appears to be

$$G_s(n)=\sum_{\substack{a_1,\ldots,a_s\geqslant 0\\ 1\cdot a_1+\ldots+s\cdot a_s=s}}\prod_{k=1}^{s}\frac{H_k^{a_k}(n)}{k^{a_k}\cdot a_k!}.$$

How would one prove it best? Can it be seen from
$$\sum_{s=0}^{\infty}G_s(n)t^s=\prod_{k=1}^{n}\Big(1-\frac{t}{k}\Big)^{-1}\tag{$*$}\label{helpeq}$$
which is easy to prove (here $G_0(n)=1$)?

REPLY [4 votes]: Indeed (thanks to darij grinberg; writing it out just to have the question answered),
$$\sum_{s=1}^{\infty}\frac{H_s(n)}{s}t^s=\sum_{k=1}^{n}\sum_{s=1}^{\infty}\frac{1}{s}\Big(\frac{t}{k}\Big)^s=-\ln\prod_{k=1}^{n}\Big(1-\frac{t}{k}\Big)$$
so the result follows from $(*)$ and "exponential formula". In turn, $(*)$ is a consequence of
$$\prod_{k=1}^{n}\Big(1-\frac{t}{k}\Big)^{-1}=\sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}\Big(1-\frac{t}{k}\Big)^{-1}$$
which is simply the partial fraction expansion of the LHS.