Motivation
I am looking into the series
$$\begin{align} \sum_{n=2}^{\infty} (\zeta(n)^{2}-1) &= 1+ \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} \\ &= 1+ \sum_{k=1}^{\infty} \bigg{(} \frac{1}{kk!} \bigg{(} \sum_{i=1}^{k} \left[ {k \atop i} \right] \zeta(i+1) - k! \bigg{)} \bigg{)}. \end{align} . $$
Therefore, I'm interested in the sum of Stirling-zeta values: $$S_{k} := \sum_{i=1}^{k} \left[ {k \atop i} \right] \zeta(i+1). $$
It appears that no alternative expressions in terms of integrals are currently known for $S_{k}$. However, there are many such expressions that are equal or similar to the reversed Stirling zeta sums:
$$T_{k} := \sum_{i=1}^{k}{k \brack i}\zeta(k + 2-i). $$
For instance, when we define $E_{k} := \int_{0}^{1} \Big{(} \frac{\ln(x)}{x-1} \Big{)}^{k} dx = \int_{0}^{1} \Big{(} - \frac{\ln(1-x)}{x} \Big{)}^k dx := F_{k}$, we have: $$E_{k} = F_{k} = k\sum_{i=0}^{k-1}\left[ k-1 \atop i\right]\zeta(k+1-i) . $$
Question
Suppose we could find a similar integral expression for the generating function $$E_{k}(x) := k\sum_{i=0}^{k-1}\left[ k-1 \atop i\right]\zeta(k+1-i) x^{i} . $$
Furthermore, define $$S_{k} (x) := \sum_{i=1}^{k}{k \brack i}\zeta(i+1)x^{i} $$
similarly. Then I thought it might be possible to transform $E_{k}(\cdot)$ into $S_{k}(\cdot)$ if one knows how to express $$f(k,i) := \frac{\left[ k \atop i\right] }{\left[ k \atop k- i\right] } $$ succinctly in closed form.
It appears some work on this has been done, but this seems to go more into approximations than into actual closed forms. So my question is:
Can $f(k,i)$ be simplified into a succinct closed form?