Using numerical analysis and tables, nth order backward difference of exponential function is obtained. Further analyzing the obtained equation yields a special identity given as \[\sum\limits_{k\, = \,0}^n {{{( - 1)}^k}\,\frac{{{{(x - \,k)}^{n\,\, - m}}}}{{(n\, - \,k - m)!\,k!}}\,} \, = \,\,\sum\limits_{k\, = \,0}^{n\, - m} {{{( - 1)}^k}\,\frac{{{{(x - \,k)}^{n\, - m}}}}{{(n\, - \,k - m)!\,k!}}\,} = \,\,1\,\,\,\,\,\,:\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\,\,\,\,\,\,\,\,\,\,\,\,\,x \in R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{n \in \,W{\rm{ }}}\\{m\, \in \,W\,\,{\rm{:}}\,\,m \le \,n}\end{array}} \right.\]This equation yields the value of negative integer factorial, zero factorial and zero to the power zero which is currently indeterminate forms along with other unknown values. With this equation, we also conclude the product of zero and infinity is unity within certain parameters.
Binomial Transform of (s, t) – Pell Sequence
