<s>[INST] How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$? How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$?
I've spent quite some time on this problem. So far, I've simplified the right-hand side to $\frac{1}{30}(n+1)[(2n+3)(3n^3)+n(n-1)]$. But then, the algebra becomes very complicated when I add $(n+1)^4$ to both sides of the inductive hypothesis.
 [/INST] $$\underbrace{1^4+2^4+3^4\cdots\ + n^4}_{\sum_{k = 1}^n k^4} = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$$
See the Faulhaber's formula and the Bernoulli number: $\text{B}_{j}(x)$
\begin{align}
\sum_{k=0}^{n} k^p = \frac{1}{p+1} \sum_{j=0}^p (-1)^j {p+1 \choose j} B_j n^{p+1-j}=\frac{\text{B}_{p+1}(n+1)-\text{B}_{p+1}(0)}{p+1}
\end{align}
In your case
\begin{align}
\sum_{k = 1}^n k^4 &= \frac{\text{B}_{4+1}(n+1)-\text{B}_{4+1}(0)}{4+1}\\
&= \frac{1}{5}\text{B}_{5}(n+1)\\
&=\frac{1}{5}\left(\frac16 (-1 - n) + \frac53 (1 + n)^3 - \frac52 (1 + n)^4 + (1 + n)^5\right)\\
&=\frac{1}{5}\left(\frac{\left(-1-n\right)}{6}+\frac{5\left(1+n\right)^3\cdot 2}{6}-\frac{5\left(1+n\right)^4\cdot 3}{6}+\frac{\left(1+n\right)^5\cdot 6}{6}\right)\\
&=\frac{1}{5}\left(\frac{6n^5+15n^4+10n^3-n}{6}\right)\\
&=\frac{6n^5+15n^4+10n^3-n}{30}\\
&= \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n \tag*{$\Box$}
\end{align}
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