The proof proceeds by induction. === Basis for the Induction === For all $n \in \Z_{\ge 3}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \left[{n \atop n - 3}\right] = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$ $P \left({3}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left[{3 \atop 0}\right]       | r = \delta_{3 0}       | c = {{Defof|Unsigned Stirling Numbers of the First 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 $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \left[{k \atop k - 3}\right] = \binom k 6 + 8 \binom {k + 1} 6 + 6 \binom {k + 2} 6$ from which it is to be shown that: :$\displaystyle \left[{ {k + 1} \atop k - 2}\right] = \binom {k + 1} 6 + 8 \binom {k + 2} 6 + 6 \binom {k + 3} 6$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \left[{k + 1 \atop k - 2}\right]       | r = k \left[{k \atop k - 2}\right] + \left[{k \atop k - 3}\right]       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = k \left({\binom n 4 + 2 \binom {n + 1} 4}\right) + \left[{k \atop k - 3}\right]       | c = Unsigned Stirling Number of the First Kind of n with n-2 }} {{eqn | r = k \left({\binom n 4 + 2 \binom {n + 1} 4}\right) + \binom k 6 + 8 \binom {k + 1} 6 + 6 \binom {k + 2} 6       | c = Induction Hypothesis }} {{eqn | r = k \left({\dfrac {k \left({k - 1}\right) \left({k - 2}\right) \left({k - 3}\right)} {4!} + 2 \dfrac {\left({k + 1}\right) k \left({k - 1}\right) \left({k - 2}\right)} {4!} }\right)       | c = {{Defof|Binomial Coefficient}} }} {{eqn | o =        | ro= +       | r = \dfrac {k \left({k - 1}\right) \left({k - 2}\right) \left({k - 3}\right) \left({k - 4}\right) \left({k - 5}\right)} {6!}       | c =  }} {{eqn | o =        | ro= +       | r = 8 \dfrac {\left({k + 1}\right) k \left({k - 1}\right) \left({k - 2}\right) \left({k - 3}\right) \left({k - 4}\right)} {6!}       | c =  }} {{eqn | o =        | ro= +       | r = 6 \dfrac {\left({k + 2}\right) \left({k + 1}\right) k \left({k - 1}\right) \left({k - 2}\right) \left({k - 3}\right)} {6!}       | c =  }} {{eqn | r = k \left({k - 1}\right) \left({k - 2}\right) \left({\dfrac {k \left({k - 3}\right) + 2 k \left({k + 1}\right)} {4!} }\right)       | c =  }} {{eqn | o =        | ro= +       | r = k \left({k - 1}\right) \left({k - 2}\right) \left({k - 3}\right) \dfrac {\left({k - 4}\right) \left({k - 5}\right) + 8 \left({k + 1}\right) \left({k - 4}\right) + 6 \left({k + 2}\right) \left({k + 1}\right)} {6!}       | c =  }} {{eqn | r = k \left({k - 1}\right) \left({k - 2}\right) \left({\dfrac {k^2 - 3 k + 2 k^2 + 2 k} {4!} }\right)       | c =  }} {{eqn | o =        | ro= +       | r = k \left({k - 1}\right) \left({k - 2}\right) \left({k - 3}\right) \dfrac {k^2 - 9 k + 20 + 8 \left({k^2 - 3 k - 4}\right) + 6 \left({k^2 + 3 k + 2}\right)} {6!}       | c =  }} {{eqn | r = k \left({k - 1}\right) \left({k - 2}\right) \left({\dfrac {3 k^2 - k} {4!} }\right)       | c =  }} {{eqn | o =        | ro= +       | r = k \left({k - 1}\right) \left({k - 2}\right) 15 \left({k - 3}\right) \dfrac {k^2 - k} {6!}       | c =  }} {{eqn | r = k \left({k - 1}\right) \left({k - 2}\right) 15 \left({\dfrac {6 k^2 - 2 k + k^3 - 4 k^2 + 3 k} {6!} }\right)       | c =  }} {{eqn | r = k \left({k - 1}\right) \left({k - 2}\right) 15 \left({\dfrac {k^3 + 2 k^2 + k} {6!} }\right)       | c =  }} {{eqn | r = \left({k + 1}\right) k \left({k - 1}\right) \left({k - 2}\right) 15 \left({\dfrac {k^2 + k} {6!} }\right)       | c =  }} {{eqn | r = \left({k + 1}\right) k \left({k - 1}\right) \left({k - 2}\right) \left({\dfrac {15 k^2 + 15 k} {6!} }\right)       | c =  }} {{eqn | r = \left({k + 1}\right) k \left({k - 1}\right) \left({k - 2}\right) \left({\dfrac {\left({k^2 - 7 k + 12}\right) + \left({8 k^2 - 8 k - 48}\right) + \left({6 k^2 + 30 k + 36}\right)} {6!} }\right)       | c =  }} {{eqn | r = \left({k + 1}\right) k \left({k - 1}\right) \left({k - 2}\right) \left({\dfrac {\left({k - 3}\right) \left({k - 4}\right) + 8 \left({k + 2}\right) \left({k - 3}\right) + 6 \left({k + 3}\right) \left({k + 2}\right)} {6!} }\right)       | c =  }} {{eqn | r = \dfrac {\left({k + 1}\right) k \left({k - 1}\right) \left({k - 2}\right) \left({k - 3}\right) \left({k - 4}\right)} {6!}       | c =  }} {{eqn | o =        | ro= +       | r = 8 \dfrac {\left({k + 2}\right) \left({k + 1}\right) k \left({k - 1}\right) \left({k - 2}\right) \left({k - 3}\right)} {6!}       | c =  }} {{eqn | o =        | ro= +       | r = 6 \dfrac {\left({k + 3}\right) \left({k + 2}\right) \left({k + 1}\right) k \left({k - 1}\right) \left({k - 2}\right)} {6!}       | c =  }} {{eqn | r = \binom {k + 1} 6 + 8 \binom {k + 2} 6 + 6 \binom {k + 3} 6       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 3}: \left[{n \atop n - 3}\right] = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$ {{qed}}