Stirling number of the first kind Identities Question: Let ${h_n}$, from n=0 to n=infinity, denote any sequence of real numbers. For n = 0,1,2, ... and using $g_n= \sum_{k=0}^n S(n, k)*h_k$, where $S(n, k)$ is a Stirling number of the second kind, and using $H_n = \sum_{k=0}^n (–1)^{n-k}*s(n, k) \;g_k$, where $s(n,k)$ is a Stirling number of the first kind, prove that $H_n = h_n$ using induction. (All sums go from k=0 to k=n). 
I am trying to prove the question above but in order to do that, I need to somehow get $s(m+1, m) = 1$ (first kind Stirling No.), is this possible? I think in the final few steps, we should have $H_{n+1} = -H_n + h_n + h_{n+1}$ and thus getting $H_{n+1}=h_{n+1}$.
If the answer to my previous question is "no", then could you please explain how I can prove this?
Thanks
 A: We seek to verify that with $H_n$ a sequence of real numbers we have
$$H_n = \sum_{k=0}^n (-1)^{n-k} {n\brack k}
\sum_{p=0}^k {k\brace p} H_p.$$
The RHS is
$$\sum_{p=0}^n H_p \sum_{k=p}^n (-1)^{n-k} {n\brack k} {k\brace p}.$$
Therefore we must show
$$[[n=p]] = \sum_{k=p}^n (-1)^{n-k} {n\brack k} {k\brace p}.$$
Using the standard EGFs the RHS becomes
$$\sum_{k=p}^n (-1)^{n-k}
n! [z^n] \frac{1}{k!} \left(\log\frac{1}{1-z}\right)^k
k! [w^k] \frac{(\exp(w)-1)^p}{p!}
\\ = (-1)^n n! [z^n] \sum_{k=p}^n (-1)^{k}
\left(\log\frac{1}{1-z}\right)^k
[w^k] \frac{(\exp(w)-1)^p}{p!}.$$
Since $\log\frac{1}{1-z} = z+\cdots$ the coefficient extractor $[z^n]$
enforces the upper limit of the sum and we get
$$(-1)^n n! [z^n] \sum_{k\ge p} (-1)^{k}
\left(\log\frac{1}{1-z}\right)^k
[w^k] \frac{(\exp(w)-1)^p}{p!}.$$
Again since $\exp(w)-1 = w+\cdots$ the coefficient extractor $[w^k]$
covers the entire exponential term in $w$ (we sum for $k\ge p$) and we
have at last
$$(-1)^n n! [z^n]
\frac{1}{p!} \left(\exp\left(-\log\frac{1}{1-z}\right)-1\right)^p
\\ = (-1)^n n! [z^n] \frac{1}{p!} ((1-z)-1)^p
= (-1)^{n+p}  n! [z^n] \frac{1}{p!} z^p.$$
This is
$$(-1)^{n+p} n! \frac{1}{p!} [[n=p]] = [[n=p]]$$
as claimed.
