An identity involving binomial coefficients and Stirling numbers of both kinds

It looks like my first response interprets the problem statement to use unsigned Stirling numbers of the first kind. We find for signed ones,

$$S_n = (-1)^{n+1} \sum_{j=n}^{2n} \sum_{k=j+1-n}^j (-1)^k 2^{j-k} {2n\choose j} {j\brace k} {k\brack j+1-n}.$$

With the usual EGFs we get

$$(-1)^{n+1} \sum_{j=n}^{2n} \sum_{k=j+1-n}^j (-1)^k 2^{j-k} {2n\choose j} j! [z^j] \frac{(\exp(z)-1)^k}{k!} \\ \times k! [w^k] \frac{1}{(j+1-n)!} \left(\log\frac{1}{1-w}\right)^{j+1-n}.$$

Now we have

$${2n\choose j} j! \frac{1}{(j+1-n)!} = \frac{(2n)!}{(2n-j)! \times (j+1-n)!} = \frac{(2n)!}{(n+1)!} {n+1\choose j+1-n}.$$

This yields for the sum

$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} \sum_{j=n}^{2n} {n+1\choose j+1-n} 2^j \\ \times [z^j] \sum_{k=j+1-n}^j (-1)^k 2^{-k} (\exp(z)-1)^k [w^k] \left(\log\frac{1}{1-w}\right)^{j+1-n} \\ = (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n \sum_{j=0}^{n} {n+1\choose j+1} 2^j \\ \times [z^{n+j}] \sum_{k=j+1}^{j+n} (-1)^k 2^{-k} (\exp(z)-1)^k [w^k] \left(\log\frac{1}{1-w}\right)^{j+1}.$$

Observe that $(\exp(z)-1)^k = z^k + \cdots$ and hence we may extend the inner sum beyond $j+n$ due to the coefficient extractor $[z^{n+j}].$ We find

$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n \sum_{j=0}^{n} {n+1\choose j+1} 2^j \\ \times [z^{n+j}] \sum_{k\ge j+1} (-1)^k 2^{-k} (\exp(z)-1)^k [w^k] \left(\log\frac{1}{1-w}\right)^{j+1}.$$

Furthermore note that $\left(\log\frac{1}{1-w}\right)^{j+1} = w^{j+1} +\cdots$ so that the coefficient extractor $[w^k]$ covers the entire series, producing

$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n \sum_{j=0}^{n} {n+1\choose j+1} 2^j [z^{n+j}] \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}.$$

Working with formal power series we are justified in writing

$$[z^{n+j}] \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1} = [z^{n-1}] \frac{1}{z^{j+1}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}$$

because the logarithmic term starts at $(-1)^{j+1} z^{j+1}/2^{j+1}.$ To see this write

$$-\frac{\exp(z)-1}{2} + \frac{1}{2} \frac{(\exp(z)-1)^2}{2^2} - \frac{1}{3} \frac{(\exp(z)-1)^3}{2^3} \pm \cdots$$

We continue

$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}] \sum_{j=0}^{n} {n+1\choose j+1} 2^{j+1} \frac{1}{z^{j+1}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1} \\ = (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}] \sum_{j=1}^{n+1} {n+1\choose j} 2^{j} \frac{1}{z^{j}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j}.$$

The term for $j=0$ in the sum is one and hence only contributes to $n=1$ so that we may write

$$-[[n=1]] + (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}] \sum_{j=0}^{n+1} {n+1\choose j} 2^{j} \frac{1}{z^{j}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j} \\ = -[[n=1]] + (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}] \left(1+\frac{2}{z} \log\frac{1}{1+(\exp(z)-1)/2}\right)^{n+1}.$$

Finally observe that

$$\left(1+\frac{2}{z} \log\frac{1}{1+(\exp(z)-1)/2}\right)^{n+1} \\ = \left(1+\frac{2}{z} \left( -\frac{\exp(z)-1}{2} + \frac{1}{2} \frac{(\exp(z)-1)^2}{2^2} - \frac{1}{3} \frac{(\exp(z)-1)^3}{2^3} \pm \cdots \right)\right)^{n+1} \\ = \left( -\frac{1}{4} z - \cdots \right)^{n+1}$$

and furthermore

$$[z^{n-1}] \left((-1)^{n+1} \frac{1}{4^{n+1}} z^{n+1} + \cdots \right) = 0$$

which is the claim.