This is a very simple question which has probably already been answered; I apologize if that is the case. We know that the following equation holds:
$$K(k)=\frac{\pi}{2}\sum\limits_{n=0}^\infty a_n^{\;2}k^{2n}\tag{1}$$
where $a_n=\frac{(2n-1)!!}{(2n)!!}=\frac{(2n)!}{4^n(n!)^2}$. Additionally, we have that:
$$K(k)^2=\frac{\pi^2}{4}\sum_{n=0}^\infty a_n^{\;3}(2kk')^{2n}\tag{2}$$
($k'=\sqrt{1-k^2}$). Generalization to $K(k)^n$ seems a natural possibility. To slightly support this, I have in the past formally derived (part of the derivation here although identity not explicitly stated) the following identity:
$$\sum_{n=0}^{\infty}\frac{(2n)!^m}{(n!)^{2m}}x^n\\=\left(\frac{2}{\pi}\right)^{m-1}\int_0^\frac{\pi}{2}...\int_0^\frac{\pi}{2}\frac{1}{\sqrt{1-x\left(2^m\prod\limits_{k=1}^{m-1}\cos\theta_k\right)^2}}\;d\theta_1...d\theta_{m-1}\tag{3}$$
(please correct me if this formula is wrong); this is an $m-2$-tuple integral of an elliptic integral. A special case of this is:
$$\sum_{n=0}^{\infty}\frac{(2n)!^3}{(n!)^{6}}x^n=\frac{4}{\pi^2}\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{1}{\sqrt{1-64x\cos^2\theta\cos^2\varphi}}\;d\theta\;d\varphi\tag{4}$$
By comparing $(2)$ and $(4)$, we have:
$$\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{1}{\sqrt{1-64(kk')^2\cos^2\theta\cos^2\varphi}}\;d\theta\;d\varphi=K(k)^2\tag{5}$$
The symmetry of $(5)$ makes me guess that the integrand can be split into a product of functions depending only on $\theta$ and $\varphi$ and turned into a product of two elliptic integrals (although for some reason I cannot see how to do it and Wolfram Alpha won't check $(5)$ for me).
If this is indeed the case it makes me wonder whether the RHS of $(3)$ may also be written as the product of $m-1$ complete elliptic functions $K(k)$ for some $k(x)$. In $(2)$, $(kk')^2=k^2-k^4$, so I wonder in particular whether the following conjecture is true:
$\forall m\in\mathbb{N}$ there exists a polynomial $\Pi_m(k)$ of order $m^2$ such that: $$K(k)^m=\left(\frac{\pi}{2}\right)^m\sum_{n=0}^\infty a_n^{\;m+1}\Pi_m(k)^n\tag{*}$$
Question: Is $(*)$ true? If so, what is $\Pi_m(k)$? If not where does the pattern break down (if I am right in thinking that there is a pattern)?