The proof proceeds by induction. === Basis for the Induction === For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition: :$\displaystyle {n \brace n - 3} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$ $\map P 3$ is the case: {{begin-eqn}} {{eqn | l = {3 \brace 0}       | r = \delta_{3 0}       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{eqn | r = 0       | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 3$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle {k \brace k - 3} = \binom {k + 2} 6 + 8 \binom {k + 1} 6 + 6 \binom k 6$ from which it is to be shown that: :$\displaystyle {k + 1 \brace k - 2} = \binom {k + 3} 6 + 8 \binom {k + 2} 6 + 6 \binom {k + 1} 6$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = {k + 1 \brace k - 2}       | r = \paren {k - 2} {k \brace k - 2} + {k \brace k - 3}       | c = {{Defof|Stirling Numbers of the Second Kind}} }} {{eqn | r = \paren {k - 2} \paren {\binom {n + 1} 4 + 2 \binom n 4} + {k \brace k - 3}       | c = Stirling Number of the Second Kind of n with n-2 }} {{eqn | r = \paren {k - 2} \paren {\binom {n + 1} 4 + 2 \binom n 4} + \binom {k + 2} 6 + 8 \binom {k + 1} 6 + 6 \binom k 6       | c = Induction Hypothesis }} {{eqn | r = \paren {k - 2} \paren {\dfrac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} } {4!} + 2 \dfrac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {4!} }       | c = {{Defof|Binomial Coefficient}} }} {{eqn | o =        | ro= +       | r = \dfrac {\paren {k + 2} \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {6!}       | c =  }} {{eqn | o =        | ro= +       | r = 8 \dfrac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} \paren {k - 4} } {6!}       | c =  }} {{eqn | o =        | ro= +       | r = 6 \dfrac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} \paren {k - 4} \paren {k - 5} } {6!}       | c =  }} {{eqn | r = k \paren {k - 1} \paren {k - 2}^2 \paren {\dfrac {\paren {k + 1} + 2 \paren {k - 3} } {4!} }       | c =  }} {{eqn | o =        | ro= +       | r = k \paren {k - 1} \paren {k - 2} \paren {k - 3} \dfrac {\paren {k + 2} \paren {k + 1} + 8 \paren {k + 1} \paren {k - 4} + 6 \paren {k - 4} \paren {k - 5} } {6!}       | c =  }} {{eqn | r = k \paren {k - 1} \paren {k - 2}^2 \paren {\dfrac {3 k - 5} {4!} }       | c =  }} {{eqn | o =        | ro= +       | r = k \paren {k - 1} \paren {k - 2} \paren {k - 3} \dfrac {k^2 + 3 k + 2 + 8 \paren {k^2 - 3 k - 4} + 6 \paren {k^2 - 9 k + 20} } {6!}       | c =  }} {{eqn | r = k \paren {k - 1} \paren {k - 2}^2 \paren {\dfrac {\paren {3 k - 5} } {4!} }       | c =  }} {{eqn | o =        | ro= +       | r = k \paren {k - 1} \paren {k - 2} 15 \paren {k - 3} \dfrac {k^2 - 5 k + 6} {6!}       | c =  }} {{eqn | r = k \paren {k - 1} \paren {k - 2}^2 15 \paren {\dfrac {6 k - 10} {6!} }       | c =  }} {{eqn | o =        | ro= +       | r = k \paren {k - 1} \paren {k - 2}^2 15 \paren {k - 3} \dfrac {k - 3} {6!}       | c =  }} {{eqn | r = k \paren {k - 1} \paren {k - 2}^2 15 \paren {\dfrac {6 k - 10 + k^2 - 6 k + 9} {6!} }       | c =  }} {{eqn | r = k \paren {k - 1} \paren {k - 2}^2 15 \paren {\dfrac {k^2 - 1} {6!} }       | c =  }} {{eqn | r = \paren {k + 1} k \paren {k - 1} \paren {k - 2} 15 \paren {\dfrac {\paren {k - 1} \paren {k - 2} } {6!} }       | c =  }} {{eqn | r = \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {\dfrac {15 k^2 - 45 k + 30} {6!} }       | c =  }} {{eqn | r = \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {\dfrac {\paren {6 k^2 - 42 k + 72} + \paren {8 k^2 - 8 k - 48} + \paren {k^2 + 5 k + 6} } {6!} }       | c =  }} {{eqn | r = \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {\dfrac {\paren {k + 3} \paren {k + 2} + 8 \paren {k + 2} \paren {k - 3} + 6 \paren {k - 3} \paren {k - 4} } {6!} }       | c =  }} {{eqn | r = \dfrac {\paren {k + 3} \paren {k + 2} \paren {k + 1} k \paren {k - 1} \paren {k - 2} } {6!}       | c =  }} {{eqn | o =        | ro= +       | r = 8 \dfrac {\paren {k + 2} \paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {6!}       | c =  }} {{eqn | o =        | ro= +       | r = 6 \dfrac {\paren {k + 1} k \paren {k - 1} \paren {k - 2} \paren {k - 3} \paren {k - 4} } {6!}       | c =  }} {{eqn | r = \binom {k + 3} 6 + 8 \binom {k + 2} 6 + 6 \binom {k + 1} 6       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 3}: {n \brace n - 3} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$ {{qed}}