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?
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 $$- (x-1)^{\underline{n-1}} \prod_{p=2}^{n+1} \frac{1}{x-p} = - \frac{x-1}{(x-n)(x-(n+1))}.$$