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Knowing that $$\ m^n= \sum_{k=1}^m \begin{pmatrix} m \\ k \end{pmatrix}k!S_n^k $$ $S_n^k$ being a Stirling number of second type (proved in the previous question), deduce that, for n in $\mathbb{N^*}$ : $$\ X^n= \sum_{k=1}^n S_n^k X(X-1)...(X-k+1) $$ I replaced m by X but i don't understand how am I supposed to make that n appears on the sum. If you could give me a hint, thanks by advance !

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You can write \begin{eqnarray*}\sum_{k=1}^n S_n^k X(X-1)...(X-k+1)&=&\sum_{k=1}^X \binom{X}{k}k!S_n^k+\underbrace{\sum_{k=X+1}^n \binom{X}{k}k!S_n^k}_{=0}=X^n \end{eqnarray*} since $\binom{X}{k}=0$ when $k>X$.

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