Denote $A:=\bigcup_{i=1}^n A_i$ and let us fix some $a\in A$. Suppose all the sets $a$ is an element of are $A_{i_1},A_{i_2},\dots,A_{i_m}, m\leq n$. Denote $$D_1(k):=-\sum_{i=1}^n|A_i|+2\sum_{i<j}|A_i\cap A_j |+...+(-1)^k2^{k-1}\sum_{i_1<i_2<...<i_k}|A_{i_1}\cap A_{i_2}\cap ...\cap A_{i_k}|\,\,\,\,\,\,\,\,\,(1)$$Note that the contribution of $a$ in the RHS of $(1)$ is exactly $$a(k):=-\binom{m}{1}+2^1\binom{m}{2}-\dots +(-1)^k2^{k-1}\binom{m}{k}$$in case of $k\le m$ and $$a(k):=-\binom{m}{1}+2^1\binom{m}{2}-\dots +(-1)^k2^{k-1}\binom{m}{m}=\frac{(1-2)^m-1}{2}$$in case $k>m$. In both cases $$a(k)\equiv \frac{(1-2)^m-1}{2}\pmod{2^{k}}$$Hence $a(k)\equiv 0\pmod{2^k}$ if $m$ is even and $a(k)\equiv -1\pmod{2^k}$ in case $m$ is odd. It yields $$D_1(k)=\sum_{a\in A} a(k)\equiv -d\pmod{2^k} $$and the result follows.