Sum of powers: $1^m+2^m+3^m+...+n^m$=? For any positive integer $n$ and $m,$ I was wondering if there is any way to get a closed formula for 
$$S(n,m)=1^m+2^m+3^m+\cdots+n^m$$
something like 
$$S(n,1)=1+2+3+\cdots+n=\frac{n(n+1)}{2}.$$ 
 A: This answer is the exact copy of Garvil's answer and so I've made this answer 'community wiki'.

Consider any natural number $r$. You have $$r^3-(r-1)^3=3r^2-3r+1.$$ 
Now telescope it: 
$$
1^3-0^3=3-3+1
$$
$$2^3-1^3=3\cdot2^2-3\cdot2+1
$$
$$\vdots
$$
$$
n^3-(n-1)^3=3n^2-3n+1
$$ Now add, and see them cancel out. You are left with $$n^3=3(1^2+2^2+\cdots+ n^2)-3(1+2+3+\cdots+n)+n$$ You must know 
$$
1+2+3+\cdots+n=\frac{n(n+1)}{2}.
$$ 
Plug it in, and you get the answer. Also, please see that this method works even for $\sum r^4,r^5,\cdots$. I have tried it out. All you need is the sum of its previous powers.
A: Read this: http://www.codeproject.com/Tips/792255/Faulhaber-made-easy.
Closed formulas are also known.
A: [Copied from an answer of mine at https://math.stackexchange.com/questions/1160658/how-to-find-general-formula-of-sum-n2/1160711#1160711]
There is an interesting method to find the above summations. 
Theorem. Let $n\geq r$. Then
$$\sum_{k=1}^{n}{k\choose r}={r\choose r}+...+{n\choose r}={k+1\choose r+1}$$
note that for $k<r$, $c(k,r)=0$.
Now, since $k^2=k+k(k-1)={k\choose 1}+2{k\choose 2}$
taking sum of both sides and using the above theorem gives
$$\sum_{k=1}^{n}k^2=\sum_{k=1}^{n}{k\choose 1}+2\sum_{k=1}^{n}{k\choose 2}={n+1\choose 2}+2{n+1\choose 3}=\frac{n(n+1)(2n+1)}{6}$$
I said this way of proof is more interesting because for the summation $\sum_{k=1}^{n}k^3$ one can write $$k^3=x{k\choose 1}+y{k\choose 2}+z{k\choose 3}$$ and then finds the coefficients $x,y,z$ by solving a system and finally by applying the above theorem gets $\sum_{k=1}^{n}k^3=\left(\frac{n(n+1)}{2}\right)^2$.
