Firstly note that the $r^{th}$ summand (without the $2^r$) is precisely #$\{(x,A{}_i{}_1,A{}_i{}_2,\ldots,A{}_i{}_r)|x \in A{}_{i{}_1} \cap A{}_{i{}_2} \cap \cdots \cap A{}_{i{}_k}\}$. Counting from the perspective of the points instead, we get that this sum is the same as $\sum_{x\in S} \binom{|A_x|}{r}$ where $A_x=\{i|x\in A_i\}$ and $S$ is the superset of the $A_i's$. what we want to show is that for each $1\leq k\leq n$ we have that $d(n)-\sum_x \binom{|A_x|}{1} +2\sum_x \binom{|A_x|}{2}-\cdots+(-1){}^k2{}^{k-1} \sum_x \binom{|A_x|}{k}$ = \[ d(n)-\sum_x\big{(} \binom{|A_x|}{1} +2\binom{|A_x|}{2}-\cdots+(-1){}^k2{}^{k-1}\binom{|A_x|}{k}\big{)} \] is divisible by $2^k$. Note that to prove this it suffices to show that $d(n)-\sum_x\big{(} \binom{|A_x|}{1} +2\binom{|A_x|}{2}-\cdots+(-1){}^n2{}^{n-1}\binom{|A_x|}{n}\big{)}$ is divisible by $2^n$. but this summand (for each $x$ ) is simply $\dfrac{(1-\theta)^{|A_x|}-1}{\theta}|_{\theta=2}$ which is $-1$ if $|A_x|$ is odd and zero if it is even. Hence the result.