Aristotle's solution is as follows. It is surprisingly easy.

Let $(a_n)$ be the sequence of powers of $d_i$ (sorted, with multiplicity). For example, if $d_1=2$ and $d_2=3$, then the sequences is: $1,1,2,3,4,8,9,16,27,\ldots$.

We want to show that every positive integer is a subsequence sum. This is equivalent to $a_{n+1} -1 \leq (a_1+\dots +a_n)$. The RHS is $\sum_{i=1}^k (d_i^{e_{i,n}}-1)/(d_i-1)$, where $e_{i,n}$ is the first power of $d_i$ that has not ocurred in the first $n$ terms. This is bounded below by $\min_i (d_i^{e_{i,n}}-1)$. However, $a_{n+1}=\min_i d_i^{e_{i,n}}$. Done.

Note, there is some ambiguity in the definition of $e_{i,n}$. In the example $d_1=2, d_2=3$, we can decide arbitrarily that $a_1$ is a power of $2$ and $a_2$ is a power of $3$, so $e_{2,1}=0$ but $e_{2,2}=1$.