Please note that, in the first sentence, I invoke the x < 3^k result which relies on x < 2^n, for which we need Baker/Rhin etc…
The above proof of no circuits relies on that result to say that parity vector 0^l 1^k ends in exactly 3^k - 1 and not just in some number congruent to it modulo 3^k.
The point of the proof is only to rely on this more general bound rather than developing a complex argument specific to circuits.