While manipulating the expression, $$n! = n(n-1)(n-2)\cdot\ldots\cdot 3\cdot 2\cdot 1,
$$ (I think it would be better not to tell how I manipulated because it might be more confusing) I ended up getting the sum of $n$ terms raised to the $p$-th power (as part of the whole expression thing), where $p = 1, 2, 3, 4, \ldots$
so then I was trying to derive an 'simpler' expression for sum of $n$ terms raised to the pth power from Faulhaber's formula (the below one)
$$
{\sum_{k=1}^{n}k^{p}={\frac{1}{p+1}}\sum_{k=0}^{p}{\binom{p+1}{k}}B_{k}n^{p-k+1}.}
$$
where $B_k$ are Bernoulli Numbers and ${\binom{p+1}{k}}$ is '$p$ choose $k$'.
If i take $n^p$ out of the summation (since $n$ and $p$ are constants), I can cancel and simplify some terms in my derivation, obtaining
$$
{\sum_{k=1}^{n}k^{p}=n^{p}{\frac{1}{p+1}}\sum_{k=0}^{p}{\binom{p+1}{k}}B_{k}n^{1-k}.}
$$
But is it possible now to evaluate the summation
$$
\sum_{k=0}^{p}{\binom{p+1}{k}}B_{k}n^{1-k}\;?
$$
Can you help me evaluating this expression or tell me how to go about it
I know that i am not being specific but i don't know how else to put up this questions
Thanks in advance !!!