How to show that: $(-1)^nn+2+\sum_{k=1}^{n}\frac{(-1)^k}{k+1}(2)_{k+1}S_{n}^{k+1}H_{k+1}=B_{n-1}$ I was observing these formulas concerning $B_n$, Bernoulli numbers
I was able  to conjecture this formula:
$$(-1)^nn+2+\sum_{k=1}^{n}\frac{(-1)^k}{k+1}(2)_{k+1}S_{n}^{k+1}H_{k+1}=B_{n-1}\tag1$$
Where $H_n$ is Harmonic numbers
$S_{n}^m$ is Stirling number of second kind
$(n)_m$ is Pochhammer
Does anyone know how to prove it?
 A: We seek to prove the identity
$$B_n = \sum_{k=0}^{n} \frac{(-1)^k H_{k+1} (k+2)!
{n+1\brace k+1}}{k+1} + (-1)^{n+1} (n+1).$$
The sum is
$$(n+1)! [z^{n+1}]
\sum_{k=0}^{n} (-1)^k \left(1+\frac{1}{k+1}\right)
H_{k+1} (\exp(z)-1)^{k+1}.$$
With $\exp(z)-1 = z+\cdots$ the coefficient extractor enforces the upper
limit of the sum  and we get
$$(n+1)! [z^{n+1}]
\sum_{k\ge 0} (-1)^k \left(1+\frac{1}{k+1}\right)
(\exp(z)-1)^{k+1} [w^{k+1}] \frac{1}{1-w} \log\frac{1}{1-w}.$$
Note that the term in $w$ starts at $w$. We get for the first piece
$$-(n+1)! [z^{n+1}] \exp(-z) \log \exp(-z)
\\ = (n+1)! [z^n] \exp(-z) = (-1)^n (n+1).$$
We see that this cancels the extra term from the initial closed form.
Therefore the remaining term must give the Bernoulli numbers:
$$(n+1)! [z^{n+1}]
\sum_{k\ge 0} (-1)^k \frac{1}{k+1}
(\exp(z)-1)^{k+1} [w^{k+1}] \frac{1}{1-w} \log\frac{1}{1-w}.$$
Differentiate to get
$$n! [z^n] \exp(z)
\sum_{k\ge 0} (-1)^k 
(\exp(z)-1)^{k} [w^{k+1}] \frac{1}{1-w} \log\frac{1}{1-w}
\\ = n! [z^n] \exp(z)
\sum_{k\ge 0} (-1)^k
(\exp(z)-1)^{k} [w^k] \frac{1}{w} \frac{1}{1-w} \log\frac{1}{1-w}.$$
This is
$$n! [z^n] \exp(z) \frac{1}{1-\exp(z)} \exp(-z) \log\exp(-z)
\\ = n! [z^n] \frac{z}{\exp(z)-1} = B_n$$
as claimed.
