For given $k,N\in\mathbb{N}$, how to compute $\sum_{i=0}^Ni^k$? For given $k$ and $N$, $k,N\in\mathbb{N}$, how to compute $\sum_{i=0}^Ni^k$?
We have: $\sum^N_{i=0}i=\frac{N(N+1)}2$
Also according to what I found in the Internet we have $\sum^N_{i=0}i^2=\frac{n(n+1)(2n+1)}{6}$ (I can prove this formula by induction, but I couldn't obtain it myself)
But how to generalize this for any $k$?
 A: We have
\begin{eqnarray*}
\sum_{i=1}^{n} 1 =n \\
\sum_{i=1}^{n} i =\frac{n(n+1)}{2} \\
\end{eqnarray*}
The following is a telescoping sum 
\begin{eqnarray*}
\sum_{i=1}^{n} [(i+1)^3-i^3] =(n+1)^3-1. \\
\end{eqnarray*}
It is now a bit of a grind but with these results you will get to 
\begin{eqnarray*}
\sum_{i=1}^{n} i^2 =\frac{n(n+1)(2n+1)}{6}. \\
\end{eqnarray*}
A similar trick will allow you to calculate higher powers, but the calculation gets longer each time.
You are probably best to do as dxiv suggests & use the reference to Faulhaber's formula.
A: Here is an induction formula from which all sum can be derived:
Let $$ S_n(p)=\sum_{k=1}^{n} k^p\qquad n, p\in\mathbb N ~~~~~\text{called Cavalieri sum of oder p}$$
then, 
We know the following Binomial formula 
$$ (k+1)^p = k^p+ \sum_{i=0}^{p-1}\binom{p}{i} k^i$$
where $\binom{p}{i}= \frac{p!}{i!(p-i)!}$.
Which implies that, 
$$\sum_{k=1}^{n} (k+1)^p =\sum_{k=1}^{n} k^p+\sum_{i=0}^{p-1}\binom{p}{i} \sum_{k=1}^{n} k^i = S_n(p) +\sum_{i=0}^{p-1}\binom{p}{i} S_n(i) $$
But $$\sum_{k=1}^{n} (k+1)^p = \sum_{k=2}^{n+1} k^p = S_{n+1}(p) -1 = S_n(p) +(n+1)^p -1$$
Hence finally we get the formula :
$$\color{red}{(n+1)^p -1  =\sum_{i=0}^{p-1}\binom{p}{i} S_n(i)} $$

From this it is possible to compute the sum for any $p\ge 1 $ in $ \mathbb N $.

