I asked this question here

When I was in high school, I was fascinated by $\displaystyle\sum\limits_{k=1}^n k= \frac{n(n+1)}{2}$ so I tried to find the general solution for $\displaystyle\sum\limits_{k=1}^n k^i$ s.t $i \in \mathbb{N}$.

I was able to to find the sum up to $i=6$. Here, I tried to search for a pattern to find the general solution of $\sum\limits_{k=1}^n k^i $ s.t $i \in \mathbb{N}$ which I failed to do, but I noticed this pattern:

$\\[3pt]$
$$S_1(n):={\sum\limits_{k=1}^n k= \frac{n(n+1)}{2}}$$

$$S_2(n):={\sum\limits_{k=1}^n k^2= \color{blue}{\frac{n(2n+1)(n+1)}{6}}}$$

$$S_3(n):={\sum\limits_{k=1}^n k^3= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}}}$$

$$S_4(n):={\sum\limits_{k=1}^n k^4=\color{blue}{\frac{n(2n+1)(n+1)}{6}} \times \frac{3n^2+3n-1}{ 5}}$$

$$S_5(n):={\sum\limits_{k=1}^n k^5= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}} \times\frac{2n^2+2n-1}{ 3}}$$

$$S_6(n):={\sum\limits_{k=1}^n k^6= \color{blue}{\frac{n(2n+1)(n+1)}{6}} \times \frac{3n^4+6n^3-3n+1}{ 7 }}$$

$$S_7(n):={\sum\limits_{k=1}^n k^7= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}} }\times\frac{3n^4+6n^3-n^2-4n+2}{ 6}}$$

$$S_8(n):={\sum\limits_{k=1}^n k^8= \color{blue}{\frac{n(2n+1)(n+1)}{6}} \times \frac{5n^6+15n^5+5n^4-15n^3-n^2+9n+3}{ 15}}$$

$$S_9(n):={\sum\limits_{k=1}^n k^9= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}} \times\frac{(n^2+n-1)(2n^4 +4n^3-n^3-3n^2+3)}{ 5}}$$

$$S_{10}(n):={\sum\limits_{k=1}^n k^{10}= \color{blue}{\frac{n(2n+1)(n+1)}{6}} \times \frac{ 3 n^8+ 12 n^7+ 8 n^6 - 18 n^5- 10 n^4+ 24 n^3 + 2 n^2 - 15 n +5}{ 11}}$$

$$S_{11}(n):={\sum\limits_{k=1}^n k^{11}= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}} \times\frac{2n^8 +8n^7+4n^6-16n^5-5n^4+26n^3-3n^2-20n+10}{ 6}}$$

$\\[3pt]$ I noticed that:

For odd $i>1$, $\displaystyle\sum\limits_{k=1}^n k^i= \color{red}{\frac{{n^2(n+1)^2}}{{4}}} \times \frac{P_{i-3}(n)}{N_i}$ s.t $P_{i-3}(n)$ is an $i-3$ polynomial with integer coefficients $ \{a_{i-3},\dots a_1,a_0 \}$ such that $\gcd \{a_{i-3},\dots a_1,a_0 \}=1$, $N_i\in \mathbb {N}$.

For even $i$, $\displaystyle\sum\limits_{k=1}^n k^i= \color{blue}{\frac{n(2n+1)(n+1)}{6}}\times \frac{P_{i-2}'(n)}{N_i}$ s.t $P_{i-2}'(n)$ is an $i-2$ polynomial with integer coefficients $\{ a_{i-2},\dots a_1,a_0 \}$ such that $\gcd \{ a_{i-2},\dots a_1,a_0 \}=1$, $N_i\in \mathbb {N}$.

When I was in high school I couldn't prove this pattern, and I remembered this observation that I had totally forgotten about. Now, after two years from my first attempt, I tried to prove this pattern again, but I couldn't.

This question has been partly answered here (The answer shows that $\displaystyle\sum\limits_{k=1}^n k^i$ is divisible by ${n^2(n+1)^2}$ for odd $i>1$ and $\displaystyle\sum\limits_{k=1}^n k^i$ is divisible by ${n(2n+1)(n+1)}$ for even $i$) the only missing part is to show that the denominator is a multiple of $4$ if $i\in 2\mathbb{N}+1$, and the denominator is a multiple of $6$ if $i \in 2\mathbb{N}$.