Using combinatorial argument to prove that $${ n\brack 3}=\frac{1}{2}\left(n-1\right)!\left[\left(H_{n-1}\right)^{2}-H_{n-1}^{\left(2\right)}\right]$$
Where ${ n\brack k}$ denotes Stirling numbers of the first kind and $ H_{n}^{\left(m\right)}$ is the generalized harmonic number.
It should be shown that the number of permutations on $[n]$ with $3$ cycles is counted by the right-hand side. Assume the first element in each one of the cycles is the least element contained in that cycle and that the cycles are ordered increasingly according to their first element (the leftmost element is considered as the first element).
$$(a_{1}...a_{j_1})(a_{j_1+1}...a_{j_2})(a_{j_2+1}...n)$$
Based on the assumption $a_1=1$ and $1\le j_{1}\le n-2$, let's call the leftmost cycle left cycle and the two other right cycles. Summing all over possible ways to distribute the remaining $n-1$ elements to three cycles gives the following summation:
$$\sum_{\;\;\;\;\;\;k_{2},k_3\ge1,\\ k_{1}+k_{2}+k_{3}=n-1}^{ }\binom{n-1}{k_1}\binom{n-1-k_1}{k_2}\binom{n-1-k_1-k_2}{k_3}(k_1)!(k_2-1)!(k_3-1)!$$
$$=\left(n-1\right)!\sum_{\;\;\;\;\;\;k_{2},k_3\ge1,\\ k_{1}+k_{2}+k_{3}=n-1}^{ }\frac{1}{k_{2}k_{3}}$$
Where $k_1$ is the left cycle and $k_2,k_3$ are the right cycles.
But I don't know how to continue, it would be appreciated if someone helps me.
Besides I like to know how much of my work is right.