Given that $$j=\frac{1}{13824q^2}\left(2^8q^2\prod_{k\gt 0}(1+q^{2k})^{16}+\prod_{k\gt 0}(1+q^{2k-1})^{16}+\prod_{k\gt 0}(1-q^{2k-1})^{16}\right)^3,$$ how can I show that $$j=\frac{1}{1728q^2}(1+c_1 q^2+c_2 q^4+\cdots)$$ where $c_1,\, c_2,\, \ldots$ are some constants? I'm interested in the first term, i.e. $\frac{1}{1728q^2}$.
I tried expanding the product, which gives $$j=\frac{1}{13824q^2}\left(2^{24}q^6\prod_{k\gt 0}(1+q^{2k})^{48}+3\cdot 2^{16}q^4\prod_{k\gt 0}(1+q^{2k})^{32}(1+q^{2k-1})^{16}+3\cdot 2^{16}q^4\prod_{k\gt 0}(1+q^{2k})^{32}(1-q^{2k-1})^{16}+3\cdot 2^{16}q^4\prod_{k\gt 0}(1+q^{2k})^{16}(1+q^{2k-1})^{32}+6\cdot 2^8q^2\prod_{k\gt 0}(1+q^{2k}-q^{4k-2}-q^{6k-2})^{16}+3\cdot 2^8 q^2\prod_{k\gt 0}(1+q^{2k})^{16}(1-q^{2k-1})^{32}+\prod_{k\gt 0}(1+q^{2k-1})^{48}+3\prod_{k\gt 0}(1+q^{2k-1})^{32}(1-q^{2k-1})^{16}+3\prod_{k\gt 0}(1+q^{2k-1})^{16}(1-q^{2k-1})^{32}+\prod_{k\gt 0}(1-q^{2k-1})^{48}\right).$$ Expanding doesn't seem to help.
But I know that the expression above can be written as $$j=\frac{1}{13824}\frac{(\theta _2 ^8(0)+\theta _3 ^8(0)+\theta _4 ^8(0))^3}{q^2\prod_{k\gt 0}(1-q^{2k})^{24}}$$ where $$\begin{align}\theta _2(0)&=2Pq^{\frac{1}{4}}\prod_{k\gt 0}(1+q^{2k})^2\\ \theta _3(0)&=P\prod_{k\gt 0}(1+q^{2k-1})^2\\ \theta _4(0)&=P\prod_{k\gt 0}(1-q^{2k-1})^2\end{align}$$ where $P=\prod_{k\gt 0}(1-q^{2k})$ and $$\theta _2 ^8(0)+\theta _3 ^8(0)+\theta _4 ^8(0)=\frac{3}{\pi ^4}(e_1 ^2+e_2 ^2+e_3 ^2).$$ The symbol $q$ is the nome $e^{\pi i\frac{\omega _1}{\omega _2}}$ and $e_1=\wp \left(\frac{\omega _1}{2}\right)$, $e_2=\wp \left(\frac{\omega _2}{2}\right)$ and $e_3 =\wp \left(-\frac{\omega _1+\omega _2}{2}\right)$ for the Weierstrass's elliptic function $\wp$.
Write this as
$$ j=\frac{1}{13824q^2}\big(A(q)+B(q)+B(-q)\big)^3 $$
Notice every coefficient of $A$ is an even number and $A$ is an even function.
Moreover, $B(q)+B(-q)$ removes all odd powers from $B$'s expansion and doubles the remaining coefficients, so the same can be said for $B(q)+B(-q)$.
Therefore, we may rewrite this as
$$ j=\frac{1}{13824q^2}\big(2C(q^2)\big)^3 $$
for some power series $C$. Cancelling out $2^3$ yields $1728$ below.