We have $\pi_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n}$ and $d_{n+1}(x) = 2^{n+1}(\pi_{n+1}(x) - \pi_n(x))$ for all $(x,n) \in \mathbb{R} \times \mathbb{N}$.
Justify
$$\forall x \in [0,1[, \forall k \in \mathbb{N}, \quad \pi_k(x) = \sum_{j=1}^{k} \frac{d_j(x)}{2^j}.$$