Let $\sigma_k(n)=\sum_{d|n}d^k$ be the generalized sum-of-divisors function. Let $S_n$ be the matrix defined by $[S_n]_{ij}=\sigma_i(j)$. I read a comment somewhere that
$$\det(S_n)=1!\cdot 2!\cdots n!.$$
Is this fact easy to prove?
For example:
$$S_3=\begin{pmatrix}1 & 3 & 4\\ 1 & 5 & 10 \\ 1 & 9 & 28 \end{pmatrix}$$
has determinant $12=1!\cdot 2! \cdot 3!$.
I'm guessing you would employ the formula:
$$\sigma_k(n)=\prod_{i}^k\frac{p_i^{(a_i+1)k}+1}{p_i^k-1},$$
where $n=p_1^{a_1}\cdots p_k^{a_k}$. Unfortunately, I don't know how to proceed after expanding the determinant. Or, maybe I should group columns with common divisors first? Or is this really dumb: $S_3$ is similar to the vandermonde matrix with entries $i^j$, which has determinant $1!2!\cdots n!$?
Yes you should group the columns in the following sense:
Hint let's take an example: $$S_3=\begin{pmatrix}1 & 1+2 & 1+3\\ 1 & 1+2^2 & 1+3^2 \\ 1 & 1+2^3 & 1+3^3 \end{pmatrix}$$
now if we made the operations $C_2\leftarrow C_2-C_1 $ and $C_3\leftarrow C_3-C_1 $ you will get: $$S_3\sim \begin{pmatrix}1 & 2 & 3\\ 1 & 2^2 & 3^2 \\ 1 & 2^3 & 3^3 \end{pmatrix}$$
In the general case you have to do : $$\begin{align}C_2&\leftarrow C_2-C_1\\ C_3&\leftarrow C_3-C_1\\ \cdots &\cdots\cdots \cdots\cdots\\ C_i&\leftarrow C_i-\sum_{d|i,d\neq i}C_d \\ \cdots &\cdots\cdots \cdots\cdots\\ C_n&\leftarrow C_n-\sum_{d|n,d\neq n}C_d \end{align}$$
and you will get the vandermonde matrix