Here is my derivation.
A very short Mathematica program for computing the zeta zeros is:
Clear[x, t, nn];
nn = 12;
t = 15;
a = Series[1/Zeta[x + t*I], {x, 0, nn}];
t*I + N[Coefficient[a, x^(nn - 1)]/Coefficient[a, x^nn]]
which for $t=15$ gives $0.5 + 14.1347i$
Tom Copeland has recorded what he calls "Coefficients of list partition transform: reciprocal of an exponential generating function (e.g.f.)." in the OEIS here: https://oeis.org/A133314 together with several links to papers.
This is the table starting:
1
[-1]
[-1, 2]
[-1, 6, -6]
[-1, 8, 6, -36, 24]
[-1, 10, 20, -60, -90, 240, -120]
[-1, 12, 30, -90, 20, -360, 480, -90, 1080, -1800, 720]
These numbers above appear to be the same as the coefficients in the power series expansion of $$\frac{1}{f(x)} \tag{1}$$:
Which is given by the Mathematica command:
Series[1/f[x], {x, 0, 6}]
or as a table:
TableForm[CoefficientList[Series[1/f[x], {x, 0, 4}], x]]
$$\begin{array}{l}
\frac{1}{f[0]} \\
-\frac{f'[0]}{f[0]^2} \\
\frac{f'[0]^2}{f[0]^3}-\frac{f''[0]}{2 f[0]^2} \\
-\frac{f'[0]^3}{f[0]^4}+\frac{f'[0] f''[0]}{f[0]^3}-\frac{f^{(3)}[0]}{6 f[0]^2} \\
\frac{24 f'[0]^4-36 f[0] f'[0]^2 f''[0]+6 f[0]^2 f''[0]^2+8 f[0]^2 f'[0] f^{(3)}[0]-f[0]^3 f^{(4)}[0]}{24 f[0]^5}
\end{array}$$
This is of course is essentially the same as repeated derivatives of $(1)$ if one discards signs and multiply with factorials.
In Mathematica for the Riemann zeta function this would be:
Clear[s];
D[1/Zeta[s], {s, 0}]
D[1/Zeta[s], {s, 1}]
D[1/Zeta[s], {s, 2}]
D[1/Zeta[s], {s, 3}]
D[1/Zeta[s], {s, 4}]
D[1/Zeta[s], {s, 5}]
Mathematica knows that the first derivative of $(1)$ is:
$$\frac{\partial \frac{1}{\zeta (s)}}{\partial s^1}=\frac{\zeta '(s)}{\zeta (s)^2}=\lim_{c\to 1} \, \left(\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}\right) \tag{2}$$
To get the second derivative we then recursively (as in repeated derivatives) nest the right hand side of $(2)$ into right hand side of $(2)$ to get:
$$\frac{\partial ^2\frac{1}{\zeta (s)}}{\partial s^2} = \frac{2 \zeta '(s)^2}{\zeta (s)^3}-\frac{\zeta ''(s)}{\zeta (s)^2}= \lim_{c\to 1} \, \left(\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (c+c+s-1-1)}}}\right) \tag{3}$$
To get the third derivative we insert the right hand side of $(3)$ into the right hand side of $(2)$ to get:
$$\frac{\partial ^3\frac{1}{\zeta (s)}}{\partial s^3} = \frac{6 \zeta '(s)^3+\zeta ^{(3)}(s) \zeta (s)^2-6 \zeta (s) \zeta '(s) \zeta ''(s)}{\zeta (s)^4} = \lim_{c\to 1} \, \left(\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (c+c+s-1-1)}}}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (c+c+s-1-1)}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+c+s-1-1)}-\frac{\zeta (c)}{\zeta (c+c+c+s-1-1-1)}}}}}\right) \tag{4}$$
and so on...
This should be possible to show with some insertion of variables into the nested derivative limits. But I don't know how to do induction to prove it. And I have not yet inserted the variables, which probably should be inserted where there are free standing integers (in this case maybe the ones in the numerators).
In Mathematica this would be:
Expand[Limit[(Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]), c -> 1]]
Expand[Limit[(Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1), c -> 1]]
Expand[Limit[(Zeta[
c]/((Zeta[c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1),
c -> 1]]
Expand[Limit[(Zeta[
c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1 + c -
1]))^-1))^-1))^-1), c -> 1]]
Now we apply the Mathematica FullSimplify
command to the expressions inside the limits:
FullSimplify[(Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1])]
FullSimplify[(Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1)]
FullSimplify[(Zeta[
c]/((Zeta[c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1)]
FullSimplify[(Zeta[
c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1 + c -
1]))^-1))^-1))^-1)]
This FullSimplify
then gives us (to my surprise) for the right hand side of $(2),(3)$ and $(4)$:
$$\zeta (c) \left(\frac{1}{\zeta (s)}-\frac{1}{\zeta (c+s-1)}\right) \tag{from RHS of 2}$$
$$\zeta (c)^2 \left(\frac{1}{\zeta (s)}-\frac{2}{\zeta (c+s-1)}+\frac{1}{\zeta (2 c+s-2)}\right) \tag{from RHS of 3}$$
$$\zeta (c)^3 \left(\frac{1}{\zeta (s)}-\frac{3}{\zeta (c+s-1)}+\frac{3}{\zeta (2 c+s-2)}-\frac{1}{\zeta (3 c+s-3)}\right) \tag{from RHS of 4}$$
$$\zeta (c)^4 \left(\frac{1}{\zeta (s)}-\frac{4}{\zeta (c+s-1)}+\frac{6}{\zeta (2 c+s-2)}-\frac{4}{\zeta (3 c+s-3)}+\frac{1}{\zeta (4 c+s-4)}\right)$$
Apparently the numerators inside the parentheses are binomial coefficients with alternating signs and the denominators with the Riemann zeta function look like multiples of natural numbers. This leads us to the conjectured form:
$$g(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-1}}{(n-1)!} \zeta (c)^{n-1} \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}$$
when including signs and factorials. Because of the special limit for derivatives this formula works only for the Riemann zeta function. The Gamma function should give something similar.
n = 1;
Limit[((-1)^(n - 1) Zeta[
c]^(n - 1) Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/(n - 1)!), c -> 1]
1/Zeta[s]
n = 2;
Limit[((-1)^(n - 1) Zeta[
c]^(n - 1) Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/(n - 1)!), c -> 1]
-(Derivative[1][Zeta][s]/Zeta[s]^2)
n = 3;
Limit[((-1)^(n - 1) Zeta[
c]^(n - 1) Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/(n - 1)!), c -> 1]
(2 Derivative[1][Zeta][s]^2 -
Zeta[s] (Zeta^[Prime][Prime])[s])/(2 Zeta[s]^3)
n = 4;
Limit[((-1)^(n - 1) Zeta[
c]^(n - 1) Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/(n - 1)!), c -> 1]