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I have found an interesting identity for n! , but my proof is slightly complicated using Bernoulli numbers. Can somebody find some simple proof of the following formula? $$(-1)^n n!=\sum_{k=2}^{n+1}(-1)^{k+1}\binom{n+1}{k}\sum_{i=1}^{k-1}i^n,\quad n\in N$$

It is interestig, that for Bernoulli numbers $b_{2n}$ we have very similar formula: $$b_{2n}=\sum_{k=2}^{2n+1}\frac{(-1)^{k+1}}{k}\binom{2n+1}{k}\sum_{i=1}^{k-1}i^{2n},\quad n\in N$$

where $b_{2n}$ can be defined from the equations: $$ b_0=1,\quad b_k=-\frac{1}{k+1}\sum_{i=0}^{k-1}\binom{k+1}{i} b_i,\quad k=1,2,3,\dots, $$

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I will use the known identity $$ \sum_{k=i+1}^{n+1}(-1)^{k+1}{n+1\choose k}=(-1)^i{n\choose i}, $$ which can be proved by induction.

Evaluating the right hand side of your equation, swapping the order of summation, and using the identity above: \begin{eqnarray*} \sum_{k=2}^{n+1}(-1)^{k+1}{n+1\choose k}\sum_{i=1}^{k-1}i^n&=&\sum_{i=1}^n i^n\sum_{k=i+1}^{n+1}(-1)^{k+1}{n+1\choose k}\\ &=&\sum_{i=1}^n i^n(-1)^i{n\choose i}. \end{eqnarray*} Let's call the value of this sum $A_n$.

We have $$ \sum_{i=0}^n (-1)^i{n\choose i} t^i=(1-t)^n, $$ so that $$ A_n=\left (t\frac{d}{dt}\right)^n(1-t)^n\bigg|_{t=1}. $$ For $k=1,2,\ldots,n$, we have an expression of the form $$ \left (t\frac{d}{dt}\right)^k(1-t)^n=\sum_{j=1}^k C_{j,k}\cdot t^j (1-t)^{n-j}, $$ where the coefficieints $C_{j,k}$ can be computed recursively using the product rule. The specific value $C_{k,k}=(-1)^k n(n-1)\ldots (n-k+1)$ can be shown by induction. Using this, we have $$ A_n=\left (t\frac{d}{dt}\right)^n(1-t)^n\bigg|_{t=1}=C_{n,n}=(-1)^n n!, $$ which is the desired result.

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Very nice, thank you very much. –  Marek Dec 16 '13 at 17:49

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