I was doing same computations for an exercise and I came up with the following relation
$$\sum_{k=1}^n\frac{1}{k}=\sum_{j=1}^n\frac{1}{(j-1)!}\sum_{i_1+\dots+i_j=n\\i_1,\dots i_j\geq1}\frac{1}{i_1i_2\dots i_j}$$
for all $n\geq1.$
Any idea how to prove it algebraically?
Small result: I'm sure that the two sums are not equal terms by terms.
We have using formal power series that the RHS is
$$\sum_{j=1}^n \frac{1}{(j-1)!} [z^n] \left(\log\frac{1}{1-z}\right)^j = [z^n] \sum_{j=1}^n \frac{1}{(j-1)!} \left(\log\frac{1}{1-z}\right)^j.$$
Now since $\log\frac{1}{1-z} = z + \cdots$ we may extend $j$ beyond $n$ without contributing to the coefficient extractor:
$$[z^n] \sum_{j\ge 1} \frac{1}{(j-1)!} \left(\log\frac{1}{1-z}\right)^j \\ = [z^n] \log\frac{1}{1-z} \sum_{j\ge 1} \frac{1}{(j-1)!} \left(\log\frac{1}{1-z}\right)^{j-1} \\ = [z^n] \log\frac{1}{1-z} \exp \log\frac{1}{1-z} = [z^n] \frac{1}{1-z} \log\frac{1}{1-z} = \sum_{k=1}^n \frac{1}{k} = H_n.$$
Remark. Restricting the proof to combinatorial methods we have for the combinatorial class $\mathcal{P}$ of permutations the specification as sets of labeled cycles, which is
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}}\mathcal{P} = \textsc{SET}(\textsc{CYC}(\mathcal{Z}))$$
Since permutations have EGF $\sum_{n\ge 0} n! z^n/n! = \frac{1}{1-z}$ we have the identity
$$\frac{1}{1-z} = \exp \log\frac{1}{1-z}$$
that was used above.