Define the $q$-analog $(a;q)_n = \prod_{k=0}^n \left(1 - aq^k\right)$.
I want to prove the identity $\frac{(q^2;q^2)_\infty}{(q;q)_\infty}=\frac{1}{(q;q^2)_\infty}$.
I viewed the LHS this way:
$\frac{1 - q^2}{1 - q} \frac{1 - q^4}{1 - q^2} \frac{1 - q^6}{1 - q^3} \frac{1 - q^8}{1 - q^4} \frac{1 - q^{10}}{1 - q^5} \frac{1 - q^{11}}{1 - q^6} \cdots$
and you can imagine getting the RHS by canceling the $k$th term on the top with the $2k+1$th term on the bottom.
Can we really cancel that way though? Since the gap between $k$ and $2k+1$ keeps getting bigger it might not be valid.