Very artificial combinatorics problem! First off, by taking $d=1$, we find that none of the residues are $1$ either. Let $n=p-1$, $g$ be an arbitrary primitive root mod $p$, and define the sequence $b_1,\ldots,b_{n-1}$ such that $a_i \equiv g^{b_i} \pmod{p}$ and $1 \leq b_i \leq n-1$. It then suffices to prove the following more general statement. Restatement wrote: Let $n$ be a positive integer and $S=\{b_1,\ldots,b_{n-1}\}$ be a multiset of nonzero residues modulo $n$. For every $d \mid n$ there are at least $d-1$ elements of $S$ which are not divisible by $d$. Prove that every residue modulo $n$ is congruent to the sum of the elements of a (possibly empty) subset of $S$. Of course, work modulo $n$. The idea is to add elements in "one at a time" in a clever fashion: I will show that we can always do this in a way such that with each additional element added, the number of possible sums increases by at least $1$. Because we have two possible sums with a single element (namely $0$ and whatever first element we choose—which is nonzero), this will finish the problem. Obviously, every previous possible sum is still possible after we start considering one more element $b_i$ (since we can just not include $b_i$). On the other hand, if no additional sums become possible, then if a sum $s$ was possible before, $s+b_i$ also should've been, and thus so should $s+2b_i,s+3b_i,\ldots$. This means that the set of previously possible sums is the union of some complete arithmetic progressions with common difference $d:=\gcd(n,b_i)$, so in fact the number of possible elements previously must have been a multiple of $n/d$, i.e. $kn/d$ for some $1 \leq k \leq d-1$—in particular, we can have a union of complete arithmetic progressions with common difference $d$ at most $d-1$ times, due to the fact that by by induction, up to this point the number of possible sums is strictly increasing. But we have at least $d-1$ choices of $b_i$ not divisible by $d$, so whenever this arithmetic sequence happens we should never be forced to pick some $b_i$ with $d \mid \gcd(n,b_i)$. This is not 100% rigorous, but it should be intuitively true. To prove this idea rigorously, we may use Hall's Marriage Lemma (!!) to obtain a perfect matching between the $n-1$ numbers $1,\ldots,n-1$, which represent the number of possible sums, and $b_1,\ldots,b_{n-1}$. We allow each integer $1 \leq k \leq n-1$ to be paired with precisely the $b_i$ which aren't divisible by $\gcd(k,n)$, of which there are at least $\gcd(k,n)-1$. Since for some $d \mid n$ we have exactly $d-1$ values of $k$ which have $d \mid \gcd(k,n)$, it is not hard to see that the condition for Hall's is satisfied (just consider each $\gcd$ independently), so a perfect matching does exist. Then just use the matched $b_i$ for each $k$, which is guaranteed to never "be a difference in an arithmetic progression"—i.e. provide an additional possible sum. So we are done. $\blacksquare$