While answering a recent question I have realized that the series $$S(n,m):=\sum_{k=n}^\infty{k \brack n}\frac{1}{k!(k-m)},\tag{1}$$ where ${k \brack n}$ are the Stirling numbers of the first kind, can be represented as a linear combination of Riemann $\zeta$-functions with rational coefficients.
Namely, the following identity holds for $0\le m<n$ : $$ S(n,m)=\frac{1}{m!}\sum_{i=0}^{m}{m \brack i}\zeta(n+1-i).\tag{2} $$ Observe that the sum of coefficients at $\zeta$-functions is equal to 1.
Surprisingly, Mathematica knows the analytical result of summation (1) only for $m=0$.
Can the identity (2) be derived directly from (1) by "elementary" means? If necessary, the result for $m=0$ can be regarded as given. An "elementary" proof of it can be found elsewhere. My attempts to apply it onto the case of non-zero $m$ have failed.
