Prove the summation involving Stirling numbers of the first kind I have been trying to prove or disprove this for 2 days now, but i don't even know where to begin.
$$
1 = \sum_{m=1}^{n} \sum_{k=1}^{n} \frac{x^{n-m}(-1)^{n-k-m} \left[\matrix {n\\n-m+1}\right]}{(x-k-1)(k-1)!(n-k)!}
$$
If you put back into stirling number form you have:
$$
1 = \sum_{m=1}^{n} \sum_{k=1}^{n} \frac{x^{n-m}(-1)^{n-k+1} s(n,n-m+1)}{(x-k-1)(k-1)!(n-k)!}
$$
Any advice on how to prove would be just as appreciated. My biggest problem is the recurrence relation that defines the Stirling numbers. Should i put the stirling numbers into its harmonic definition or the explicit formula both presented on the Wikipedia page?
 A: Suppose we are trying to evaluate
$$\sum_{m=1}^n \sum_{k=1}^n
\frac{x^{n-m} (-1)^{n-k-m} \left[ n \atop n-m+1 \right]}
{(x-k-1)(k-1)!(n-k)!}$$
which is
$$\sum_{k=1}^n \frac{(-1)^{n-k}}{(x-k-1)(k-1)!(n-k)!}
\times \sum_{m=1}^n (-1)^{-m} x^{n-m}
\left[ n \atop n-m+1 \right].$$
The second factor is
$$\sum_{m=0}^{n-1} (-1)^{n-m} x^m
\left[ n \atop m+1 \right]
= \sum_{m=1}^n (-1)^{n-m+1} x^{m-1}
\left[ n \atop m \right]
= \sum_{m=0}^n (-1)^{n-m+1} x^{m-1}
\left[ n \atop m \right]
\\ = -\frac{1}{x}
\sum_{m=0}^n (-1)^{n-m} x^m
\left[ n \atop m \right]
= -\frac{1}{x} x^{\underline{n}}
= - (x-1)^{\underline{n-1}}.$$
Return to the first sum and interpret it as a function in $x$ to get
$$\mathrm{Res}
\left(\sum_{k=1}^n \frac{(-1)^{n-k}}{(x-k-1)(k-1)!(n-k)!};
x=q\right) = \frac{(-1)^{n-q+1}}{(q-2)!(n-q+1)!}$$
for $2\le q\le n+1.$
Observe that we also have
$$\mathrm{Res}
\left(\prod_{p=2}^{n+1} \frac{1}{x-p}; x=q\right)
= \frac{(-1)^{n-q+1}}{(q-2)!(n-q+1)!}$$
for $2\le q\le n+1.$

Since the two functions are both  rational and the poles are simple and equal in
both cases with equal residues we have equality, so that
$$\sum_{k=1}^n \frac{(-1)^{n-k}}{(x-k-1)(k-1)!(n-k)!}
= \prod_{p=2}^{n+1} \frac{1}{x-p}.$$

Finally combine the two factors from the beginning to get
$$- (x-1)^{\underline{n-1}} \prod_{p=2}^{n+1} \frac{1}{x-p}
= - \frac{x-1}{(x-n)(x-(n+1))}.$$
