Bernoulli numbers and power & natural numbers:
No way, now back back okay. . .
Set $ S_n $ is a natural number and power, there is
$S_n=\frac{1}{k+1}*\sum\limits_{i=0}^{k} C_{k+1}^{i} * B_{i} * n^{k+1-i}$.
There are two Bernoulli numbers, the difference is $ B [1] $ positive and negative.
1>$B[1]=\frac{1}{2}$时，$S_n= \sum\limits_{i=1}^{n} i^{k}$
2>else，$S_n= \sum\limits_{i=0}^{n-1}i^{k}$
Properties of Bernoulli number, this time $ B [1] = - \ frac {1} {2} $.
$\sum\limits_{i=0}^{n}C_{n+1}^{i}*B_{i}=0$
Into
$\sum\limits_{i=0}^{n}{C_{n+1}^{i}*B_{i}}=-(n+1)*B_{i}$
Then you can open the factorial partition $ FFT $ spicy, is seeking Bernoulli's law.