The total differential squares to zero

Fix $n$ and $p+q=n$. Substitute the defining formula for $d$ twice and distribute composition over addition. This gives $d_{n-1}{d}_n{\iota }_{p,q}^n={\iota }_{p-2,q}^{n-2}{h}_{p-1,q}{h}_{p,q}+{\iota }_{p-1,q-1}^{n-2}(v_{p-1,q}{h}_{p,q}+h_{p,q-1}{v}_{p,q})+{\iota }_{p,q-2}^{n-2}{v}_{p,q-1}{v}_{p,q}$. Each term is an arrow from $C_{p,q}$ to $T_{n-2}$.

The first and last composites vanish by the two square-zero axioms; the middle parenthesis vanishes by anticommutation. Therefore $d_{n-1}{d}_n{\iota }_{p,q}^n=0$ for every $p+q=n$, including when any of the source or target components is zero.

The zero arrow $T_n\to T_{n-2}$ has these same composites with every injection. Coproduct uniqueness therefore gives $d_{n-1}{d}_n=0$. Since $n$ was arbitrary, all chain identities hold. The argument also covers an all-zero or a single-supported diagonal and uses no exactness of infinite coproducts or representative selections.