Gamma function arises when we consecutively differentiate an [Appell sequence][1]. An example of Appell polynomials are Bernoulli polynomials. When we differentiate it, the factors combine with themselves:

$$B_n'(x)=nB_{n-1}(x)$$

$$B_n''(x)=n(n-1)B_{n-2}(x)$$

$$B_n'''(x)=n(n-1)(n-2)B_{n-3}(x)$$

They are just another name for Hurwitz Zeta function:

$$B_n(x) = -n \zeta(1-n,x)$$

Thus, for $f(s,q)=\zeta(s,-q)$

$$\frac\partial{\partial q}f(s,q)= s f(s+1,q)$$

$$\frac{\partial^2}{\partial q^2}f(s,q)= s(s+1) f(s+2,q)$$


$$\frac{\partial^3}{\partial q^3}f(s,q)= s(s+1)(s+2) f(s+3,q)$$

Since Reihmann zeta is Hurwitz zeta evaluated at $q=1$, the expression you give is apparently consecutive derivative of Hurwitz Zeta, with factor $\pi^{-s}$ appearing if we normalize Hurwitz Zeta by stretching it horizontally by factor of pi.

Consecutive derivatives of Hurwitz Zeta in turn are nothing more than just polygamma function.

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For instance, here is the function $-1/x$:

![enter image description here][2]

If we add infinitely many similar functions with a shift of pi/2 each in both directions, we get $\tan x$. But if we do the same only in one direction, we get "incomplete tangent":

![http://storage7.static.itmages.ru/i/14/0910/h_1410326921_7988832_91f3fd7d7d.png][3]

The yellow one is $\operatorname{pg}(x)=\frac 1\pi \psi (\frac x\pi)$, the blue one is $\operatorname{cpg}(x)=-\frac 1\pi \psi (1-\frac x\pi)$. They obey $\operatorname{cpg}(x)+\operatorname{pg}(x)=-\cot(x)$.

Now if we differentiate cpg(x) we get:

$$(\operatorname{cpg}(x))^{(s-1)}=\pi^{-s}\Gamma(s)\zeta(s,1-\frac x\pi)$$

Compare it with yours formula:

$$\xi(2s) = \pi^{-s}\Gamma\left(s\right)\zeta(2s)$$


  [1]: https://en.wikipedia.org/wiki/Appell_sequence
  [2]: http://storage6.static.itmages.ru/i/14/0910/h_1410327935_9959844_7e87ae4078.png
  [3]: https://i.sstatic.net/XR4Is.png