Theorem. $\sum\limits_{k=1}^n k^p$ can always be written in the form $\frac{n^{p+1}}{p+1} + An^p + Bn^{p-1} + Cn^{p-2} + \cdots$
Proof. We will prove by complete induction on $p$. For the base case, notice that the theorem is true for $p = 1$, since $\sum\limits_{k=1}^n k = \frac{n(n+1)}{2} = \frac{1}{2}n^2 + \frac{1}{2}n.$ Next, suppose the theorem is true for all natural numbers less than or equal to $p$. To begin, we expand the binomial $(k+1)^{p+2}$ using the binomial theorem. That is, \begin{equation} \begin{aligned} (k+1)^{p+2} = k^{p+2} + \binom{p+2}{1}k^{p+1} + \binom{p+2}{2}k^{p} + \cdots + \binom{p+2}{p+1}k + 1 \\[3pt] (k+1)^{p+2} - k^{p+2} = \binom{p+2}{1}k^{p+1} + \binom{p+2}{2}k^{p} + \cdots + \binom{p+2}{p+1}k + 1. \end{aligned} \end{equation} Adding for $k = 1, 2, \ldots, n$, we have \begin{equation} \begin{aligned} (n+1)^{p+2} - 1 &= \sum_{k=1}^{n}\binom{p+2}{1}k^{p+1} + \sum_{k=1}^{n}\binom{p+2}{2}k^{p} + \cdots + \sum_{k=1}^{n}\binom{p+2}{p+1}k + n \\[3pt] (n+1)^{p+2} &= \binom{p+2}{1}\sum_{k=1}^{n}k^{p+1} + \binom{p+2}{2}\sum_{k=1}^{n}k^{p} + \cdots + \binom{p+2}{p+1}\sum_{k=1}^{n}k + n + 1 \\[3pt] (n+1)^{p+2} &= \frac{(p+2)!}{(p+1)!}\sum_{k=1}^{n}k^{p+1} + \frac{(p+2)!}{2!p!}\sum_{k=1}^{n}k^{p} + \cdots + \frac{(p+2)!}{(p+1)!}\sum_{k=1}^{n}k + n + 1 \\[3pt] \frac{(n+1)^{p+2}}{p+2} &= \sum_{k=1}^{n}k^{p+1} + \frac{p+1}{2}\sum_{k=1}^{n}k^{p} + \cdots + \sum_{k=1}^{n}k + \frac{n+1}{p+2} \\[3pt] \sum_{k=1}^{n}k^{p+1} &= \frac{(n+1)^{p+2}}{p+2} - \frac{p+1}{2}\sum_{k=1}^{n}k^{p} - \cdots - \sum_{k=1}^{n}k - \frac{n+1}{p+2}. \\[3pt] \end{aligned} \end{equation} For the final step, I will add the terms of the series, and generalize the coefficients. $$\sum_{k=1}^{n}k^{p+1} = \frac{(n+1)^{p+2}}{p+2} + An^{p+1} + Bn^p + Cn^{p-1} + Dn^{p-2} + \cdots$$ We have proven $Q(p+1)$. The induction step is thus complete, and we can conclude that the theorem is true for all natural numbers $p$. $\Box$
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