INTRO
Let $d\in{\bf N}$ and $n\in{\bf N}^+$ and denote by $S_d(n)$ the sum of the $d$-powers of integers $1$ to $n$: $$S_d(n)=\sum_{k=1}^nk^d$$
A simple use of the binomial theorem in developping the quantity $(k+1)^{d+2}-k^{d+2}$ and summing it yields the following recursive relation:
$$S_{d+1}(n)=\frac{1}{d+1}(\{(n+1)^{d+2}-1-\sum_{k=0}^d\binom{d+2}{k}S_k(n)).$$
Here are some examples:
$\small S_8(n)=\displaystyle\frac{n(n + 1)(2n + 1)(5n^6 + 15n^5 + 5n^4 - 15n^3 - n^2 + 9n - 3)}{90}$
$\small S_9(n)=\displaystyle\frac{n^2(n + 1)^2(n^2 + n - 1)(2n^4 + 4n^3 - n^2 - 3n + 3)}{20}$
$\small S_{10}(n)=\displaystyle\frac{n(n + 1)(2n + 1)(n^2 + n - 1)(3n^6 + 9n^5 + 2n^4 - 11n^3 + 3n^2 + 10n - 5)}{66}$
$\small S_{11}(n)=\displaystyle\frac{n^2(n + 1)^2(2n^8 + 8n^7 + 4n^6 - 16n^5 - 5n^4 + 26n^3 - 3n^2 - 20n + 10)}{24}$
Amongst all things, there are a few of interest:
Every $S_d(n)$ is a rational polynomial in $n$, where OEIS:A064538 appears.
The roots of the polynomials seem to follow some interesting pattern, as described on this thread.
For odd values of $d$, $S_d$ is divisible by $\left(\frac{n(n+1)}{2}\right)^2$. For even values of $d$, it is divisible by $\frac{n(n+1)(2n+1)}{6}$.
CONJECTURE
It seems that for odd primes $p$ we always have
$$S_p(n)=\sum_{k=1}^nk^p = n^a (n+1)^b F_p(n)$$
For some integer $a,b$ and $F_p(n)$ is always an irreducible polynomial. (irreducible over the rationals)
Is this always true ?
