How do I get the ratio of stirling numbers?

Question

How do I get the ratio: $\sum_{i=k}^n (-1)^{i+k}\left\{\begin{array}{l} n \\ i \end{array}\right\}\left[\begin{array}{c} i \\ k \end{array}\right]=\delta_{n,k}$ ? We know the ratios $x^{\underline{n}}=\sum_{k=0}^n (-1)^{n+k}\left[\begin{array}{c} n \\ k \end{array}\right]x^k$ $x^n=\sum_{k=0}^n \left\{\begin{array}{l} n \\ k \end{array}\right\}x^{\underline{k}}$
We substitute the first into the second and get $x^n=\sum_{k=0}^n \left\{\begin{array}{l} n \\ k \end{array}\right\}x^{\underline{k}}=\sum_{k=0}^n \left\{\begin{array}{l} n \\ k \end{array}\right\}\sum_{m=0}^k (-1)^{m+k}\left[\begin{array}{c} k \\ m \end{array}\right]x^m$
How do I change the summation to equate the coefficients for $x^i$?

Answer 1

First notice that the inner sum can be extended to $n$ because the Stirling number will just be zero and then notice that you can exchange the sums because they are finite. To finish, notice that $k<m$ vanishes and so $$\sum _{k=0}^n{n\brace k}\sum _{m=0}^n(-1)^{m+k}{k\brack m}x^m=\sum _{m=0}^nx^m\sum _{k=0}^n(-1)^{k+m}{n\brace k}{k\brack m}=\sum _{m=0}^nx^m\sum _{k=m}^n(-1)^{k+m}{n\brace k}{k\brack m}.$$