Simple proof of some formula for n! 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:
$$
b_0=1,\quad b_k=-\frac{1}{k+1}\sum_{i=0}^{k-1}\binom{k+1}{i} b_i,\quad k=1,2,3,\dots,
$$
 A: 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.
