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.

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 $C$ )
