Hint on power sum coefficients Please do not give anything more than a tiny hint for this question.
I know that there is a well-known formula for $$\sum_{i=1}^n i^k,$$ where $k$ is any non-negative integer. I have been able to prove that in fact it is a polynomial in $n$,
$$\sum_{i=1}^n i^k = \sum_{j=0}^{k+1} a_j n^j,$$
with high-order term $\frac 1 {k+1} n^{k+1}$ and zero constant term. In the process, I found a rather awkward method of calculating the rest of the coefficients. I'm now trying to figure out what the rest of them are. So far, I've gotten
$$a_j = \frac{k!}{j!(k-j+1)!}-\sum_{m=j+1}^{k+1}a_{m}\frac{m!}{j!(m-j+1)!},$$
where $a_j$ is the coefficient of $n^j$ (for $j\le k$). Can someone give me a tiny hint on how to proceed? Please do not go and tell me what the coefficients are, or how the rest of the proof goes, or anything like that.
 A: One hint is that it is far easier to do this is you replace $i^k$ by another polynomial of degree $k$ in $i$, namely by$~\binom ik$. Check that you can find $\sum_{i=0}^n\binom ik$ easily. Then it is theoretically only a question of transforming the basis $[1,i,i^2,\ldots]$ of the polynomial functions in$~i$ to the basis $[\binom i0=1,\binom i1=i,\binom i2=\frac{i(i-1)}2,\ldots]$ and back. In practice this messes the concrete values up considerably.
A: Look at the Riemann and Hurwitz zeta functions.
A: Please take this not yet as an answer, only it is too long for the comment-box 
Hmm, I thought to help, but I'm not sure that I understand your question at all. If I read your equation correctly, then a coeffcient $a_j$ is explained by the higher ones $a_{j+1}, a_{j+2} ,... a_{k+1}$. So this ( for a certain $k$ ) looks like a matrix equation of the following type $  C*A=A$ where $C$ contains your binomial-coefficients:
$$ \begin{array}{lll}
\begin{array}{lll}
 \end{array}&
*&\begin{bmatrix}
  1\\ a_1\\a_2\\a_3\\a_4\\a_5
 \end{bmatrix} \\
\begin{bmatrix}
 c_{1,0}&.&c_{1,2}&c_{1,3}&c_{1,4}&c_{1,5}\\
 c_{2,0}&.&. & c_{2,3}& c_{2,4}& c_{2,5}\\
 c_{3,0}&.&. &. &c_{3,4}&c_{3,5}\\
 .&.&. &. &  1 & .\\
 .&.&. &. & .& 1\\
 \end{bmatrix}&
=&\begin{bmatrix}
 a_1\\a_2\\a_3\\a_4\\a_5\\
 \end{bmatrix} \\ \end{array}$$
but where for instance $a_4$ and $a_5$ are given(?) and are inserted in the upper vector $A$ (and then "computed" into the lower vector $A$).
The iteration begins then that with them and your given coefficents $c_{j,m}$ from there $a_3$ is computed in the lower $A$-vector (and then also inserted in the upper $A$-vector) and then $a_2$ and finally $a_1$.     
So for instance , your equation 
$$a_j = \frac{k!}{j!(k-j+1)!}-\sum_{m=j+1}^{k+1}a_{m}\frac{m!}{j!(m-j+1)!}$$
for j=3 and k+1=5 becomes
$$a_3 = \frac{4!}{3!2!}-(a_4\frac{4!}{3!(2)!} + a_5\frac{5!}{3!(3)!}) $$
and this is expressed in the matrix-form by the third row where we get 
$$  a_3 = c_{3,0}+ a_4 c_{3,4} + a_5 c_{3,5} $$
(only where I have made the minus-signs in the coefficients $c_{j,m}$ to make the appearance clearer)
and then $a_2$ can be computed by the second row of the matrix-schema:
$$  a_2 = c_{2,0}+ a_3 c_{2,3}+ a_4 c_{2,4} + a_5 c_{2,5} $$
Is that correct so far?
(The first idea would then be that this can be handled as an eigenvalue/eigenvector -problem of the matrix $C$ )
