The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ {n \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{n - k} = \delta_{m n}$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \sum_k \left\{ {0 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{0 - k}       | r = \sum_k \delta_{k 0} \left[{k \atop m}\right] \left({-1}\right)^{-k}       | c = Definition of Stirling Numbers of the Second Kind }} {{eqn | r = \left[{0 \atop m}\right]       | c = as all other terms vanish by $\delta_{k 0}$ }} {{eqn | r = \delta_{m 0}       | c = Definition of Unsigned Stirling Numbers of the First Kind }} {{end-eqn}} Thus $P \left({0}\right)$ has been shown to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({r}\right)$ is true for all $r \ge 0$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ {r \atop k}\right\}\left[{k \atop m}\right]  \left({-1}\right)^{r - k} = \delta_{m r}$ from which it is to be shown that: :$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ { {r + 1} \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k} = \delta_{m \left({r + 1}\right)}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | r = \sum_k \left\{ { {r + 1} \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}       | o =        | c =  }} {{eqn | r = \sum_k \left({k \left\{ {r \atop k}\right\} + \left\{ {r \atop k - 1}\right\} }\right) \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}       | c = Definition of Stirling Numbers of the Second Kind }} {{eqn | r = \sum_k k \left\{ {r \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k} + \sum_k \left\{ {r \atop k - 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}       | c =  }} {{eqn | r = \sum_k k \left\{ {r \atop k}\right\} \frac 1 k \left({\left[{k + 1 \atop m}\right] - \left[{k \atop m - 1}\right]}\right) \left({-1}\right)^{r + 1 - k}       | c = Definition of Unsigned Stirling Numbers of the First Kind }} {{eqn | o =        | r = + \sum_k \left\{ {r \atop k - 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}       | c =  }} {{eqn | r = \sum_k \left\{ {r \atop k}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{r + 1 - k} - \sum_k \left\{ {r \atop k}\right\} \left[{k \atop m - 1}\right] \left({-1}\right)^{r + 1 - k}       | c =  }} {{eqn | o =        | r = + \sum_k \left\{ {r \atop k - 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}       | c =  }} {{eqn | r = \sum_k \left\{ {r \atop k}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{r + 1 - k} - \sum_k \left\{ {r \atop k}\right\} \left[{k \atop m - 1}\right] \left({-1}\right)^{r + 1 - k}       | c =  }} {{eqn | o =        | r = + \sum_k \left\{ {r \atop k}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{r + 1 - k + 1}       | c = Translation of Index Variable of Summation }} {{eqn | r = \sum_k \left\{ {r \atop k}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{r + 1 - k} - \sum_k \left\{ {r \atop k}\right\} \left[{k \atop m - 1}\right] \left({-1}\right)^{r + 1 - k}       | c =  }} {{eqn | o =        | r = - \sum_k \left\{ {r \atop k}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{r + 1 - k}       | c =  }} {{eqn | r = \sum_k \left\{ {r \atop k}\right\} \left[{k \atop m - 1}\right] \left({-1}\right)^{r - k}       | c = Simplification }} {{eqn | r = \delta_{\left({m - 1}\right) r}       | c = Induction Hypothesis }} {{eqn | r = \delta_{m \left({r + 1}\right)}       | c = Definition of Kronecker Delta }} {{end-eqn}} So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left\{ {n \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{n - k} = \delta_{m n}$ {{qed}}