$\newcommand\rk{\operatorname{rk}}$Let $\Delta_n$ denote the full $n$-simplex $\{n,\dotsc,0\}$. It is clear that there are $\binom{n+1}{d+1}$ many $d$-simplices, since a $d$-simplex corresponds to a subset of $\Delta_n$ of $d+1$ elements. This means that the cochains $C_d(\Delta_n)$ are free of rank $\binom{n+1}{d+1}$.
I would like to know the ranks of the cycles $Z_d(\Delta_n)$ and the boundaries $B_d(\Delta_n)$. Certainly we have $\rk C_d(\Delta_n) = \rk B_{d-1}(\Delta_n) + \rk Z_d(\Delta_n)$, and since $\Delta_n$ is contractible, we get $\rk Z_{d}(\Delta_n) = \rk B_d(\Delta_n)$.
I have read that $\rk Z_d(\Delta_n) = \binom{n}{d+1}$; the above formula then reflects the recursion formula for binomial coefficients. However, I struggle to see how the rank of $\rk Z_d(\Delta_n)$ is obtained.
As mentioned in the comments, the correct equation is $\text{rk } Z_d(\Delta^n)=\binom{n}{d+1}$.
Well, using the two equations you gave, you can prove that $\text{rk } Z_d(\Delta^n)=\binom{n}{d+1}$ by induction on $d$. For simplicity, let $z_d=\text{rk }Z_d(\Delta^n)$, and $b_d=\text{rk } B_d(\Delta^n)$. \begin{align} z_d &=\binom{n+1}{d+1}-b_{d-1}\tag{rank-nullity}\\ &=\binom{n+1}{d+1}-z_{d-1}\tag{contractible}\\ &=\binom{n+1}{d+1}-\binom{n}{d}\tag{induction hypothesis}\\ &=\binom{n}{d+1} \end{align}