The proof proceeds by induction on $n$. For all $n \in \Z_{\ge 0}$, let $P \left({m}\right)$ be the proposition: :$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ {n + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} = \binom n m$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | r = \sum_k \left\{ {0 + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}       | o =        | c =  }} {{eqn | r = \sum_k \delta_{1 \left({k + 1}\right)} \left[{k \atop m}\right] \left({-1}\right)^{k - m}       | c = Stirling Number of the Second Kind of 1 }} {{eqn | r = \sum_k \delta_{0 k} \left[{k \atop m}\right] \left({-1}\right)^{k - m}       | c = {{Defof|Kronecker Delta}} }} {{eqn | r = \left[{0 \atop m}\right] \left({-1}\right)^{- m}       | c = All terms but where $k = 0$ vanish }} {{eqn | r = \delta_{0 m} \left({-1}\right)^{- m}       | c = Unsigned Stirling Number of the First Kind of 0 }} {{eqn | r = \delta_{0 m}       | c = multiplier irrelevant when $m \ne 0$ }} {{eqn | r = \binom 0 m       | c = Zero Choose n }} {{end-eqn}} So $P \left({0}\right)$ is seen 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, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \sum_k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} = \binom r m$ from which it is to be shown that: :$\displaystyle \sum_k \left\{ {r + 2 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} = \binom {r + 1} m$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | o =        | r = \sum_k \left\{ {r + 2 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}       | c =  }} {{eqn | r = \sum_k \left({\left({k + 1}\right) \left\{ {r + 1 \atop k + 1}\right\} + \left\{ {r + 1 \atop k}\right\} }\right) \left[{k \atop m}\right] \left({-1}\right)^{k - m}       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{eqn | r = \sum_k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}       | c =  }} {{eqn | o =        | ro= +       | r = \sum_k k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} + \sum_k \left\{ {r + 1 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}       | c =  }} {{eqn | r = \binom r m + \sum_k k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} + \sum_k \left\{ {r + 1 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}       | c = Induction Hypothesis }} {{eqn | r = \binom r m + \sum_k k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}       | c =  }} {{eqn | o =        | ro= +       | r = \left({-1}\right)^{r + 1 - m} \sum_k \left\{ {r + 1 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}       | c =  }} {{eqn | r = \binom r m + \sum_k k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} +  \left({-1}\right)^{r + 1 - m} \delta_{m \left({r + 1}\right)}       | c = Second Inversion Formula for Stirling Numbers }} {{eqn | r = \sum_k k \left\{ {r + 1 \atop k + 1}\right\} \left({\frac 1 k \left[{k + 1 \atop m}\right] - \frac 1 k \left[{k \atop m - 1}\right]}\right) \left({-1}\right)^{k - m}       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | o =        | ro= +       | r = \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}       | c =  }} {{eqn | r = \sum_k \left\{ {r + 1 \atop k + 1}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{k - m} + \sum_k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m - 1}\right] \left({-1}\right)^{k - m}       | c =  }} {{eqn | o =        | ro= +       | r = \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}       | c =  }} {{eqn | r = \sum_k \left\{ {r + 1 \atop k + 1}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{k - m} + \binom r {m - 1}       | c = Induction Hypothesis }} {{eqn | o =        | ro= +       | r = \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}       | c =  }} {{eqn | r = \sum_k \left\{ {r + 1 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - 1 - m} + \binom r {m - 1}       | c = Translation of Index Variable of Summation }} {{eqn | o =        | ro= +       | r = \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}       | c =  }} {{eqn | r = \left({-1}\right)^{r - m} \sum_k \left\{ {r + 1 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k} + \binom r {m - 1}       | c =  }} {{eqn | o =        | ro= +       | r = \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}       | c =  }} {{eqn | r = \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)} + \binom r {m - 1} + \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}       | c = Second Inversion Formula for Stirling Numbers }} {{eqn | r = \binom r {m - 1} + \binom r m       | c = simplifying }} {{eqn | r = \binom {r + 1} m       | c = Pascal's Rule }} {{end-eqn}} So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left\{ {n + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} = \binom n m$ {{qed}}