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