We revisit the q-deformed counterpart of the Zassenhaus formula, expressing
the Jackson $q$-exponential of the sum of two non-$q$-commuting operators as an
(in general) infinite product of $q$-exponential operators involving repeated
$q$-commutators of increasing order, $E_q(A+B) = E_{q^{\alpha_0}}(A)
E_{q^{\alpha_1}}(B) \prod_{i=2}^{\infty} E_{q^{\alpha_i}}(C_i)$. By
systematically transforming the $q$-exponentials into exponentials of series
and using the conventional Baker-Campbell-Hausdorff formula, we prove that one
can make any choice for the bases $q^{\alpha_i}$, $i=0$, 1, 2, ..., of the
$q$-exponentials in the infinite product. An explicit calculation of the
operators $C_i$ in the successive factors, carried out up to sixth order, also
shows that the simplest $q$-Zassenhaus formula is obtained for $\alpha_0 =
\alpha_1 = 1$, $\alpha_2 = 2$, and $\alpha_3 = 3$. This confirms and reinforces
a result of Sridhar and Jagannathan, based on fourth-order calculations.