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