# Proof that every sum of exponents can be represented as a polynomial. I am missing an inital idea.

$$s_n(p)=\sum_{k=1}^n k^p$$

Show: For every $$q \geq 1$$ exist rational numbers $$a_{k,q} , 1 \leq k \leq q-1$$, such that

$$s_n(q)= \frac 1 {q+1} n^{q+1}+ \frac 1 2 n^q + \sum_{k=1}^{q-1} a_{k,q}n^{q-k}$$

I am afraid I don't have a clue on how to prove that. Induction? Rearranging? Before that task the Pascal Identity was introduced:

$$\sum_{p=0}^q \binom{q+1}p s_n(p)=(n+1)^{q+1} - 1$$

But I can't really use that, because $$s_n(p)$$ refers to p and not to q, doesn't it?

Have you got any ideas on how to prove that?

• Hint: Put the Binomial Theorem expansion of $(k+1)^{p+1}$ into $\frac {1}{p+1}((k+1)^{p+1}-k^{p+1})$ and then sum over $k=1,...n.$ Then use induction on $p.$ – DanielWainfleet Apr 19 at 12:19

Prove this is true for $$s_n(0)$$, which should be easy. Then, given $$q$$, write $$(q+1)s_n(q)=-\sum_{p=0}^{q-1} \binom{q+1}p s_n(p)+(n+1)^{q+1} - 1$$ There are a few more nontrivial steps to complete from here, but this at least tells you it is a polynomial in $$n$$.