Combinatorial Interpretation of the Sum-of-Divisor Function?

Question

A problem in my discrete mathematics course was to prove combinatorially that $d(n) = \sum_{d|n}1 = \prod_{k = 1}^r (\alpha_k + 1)$ for a positive integer $n = \prod_{k = 1}^rp_k^{\alpha_k}$, where $p_k$ is a prime factor of $k$, and $\alpha_k$ a positive integer. Is there a combinatorial interpretation of $\sigma(n)$ (the sum-of-divisor function) that may be used to prove $\sigma(n) = \prod_{k = 1}^r \frac{p_k^{\alpha_k + 1} - 1}{p_k - 1}$?

Answer 1

I guess the shown combinatorial interpretation for $d(n)=\prod_{p\mid n}\left(1+\nu_p(n)\right)$ has been

Lemma. $d\mid n$ iff the prime divisors of $d$ are a subset of the prime divisors of $n$, and for each prime $p\mid d$ we have $\nu_p(d)\leq \nu_p(n)$.

In other terms, $d(n)$ is the cardinality of a cartesian product of $\omega(n)$ finite sets.
Similarly, if we have $E=A_1\times\ldots\times A_k$ then $$ \sum_{(a_1,\ldots,a_k)\in E}\prod_{j=1}^{k}a_j=\prod_{j=1}^{k}\sum_{a\in A_j}a $$ so $\sigma(n)$, which is a multiplicative function (in terms of Dirichet convolutions, $d=\mathbb{1}*\mathbb{1}$ implies $\sigma=\mathbb{1}*\mathbb{1}*\mathbb{1}$), clearly equals $$ \prod_{j=1}^{\omega(n)}\sum_{k=0}^{\nu_{p_j}(n)}p_j^k = \prod_{j=1}^{\omega(n)}\frac{p^{\nu_{p_j}(n)+1}-1}{p_j-1}.$$