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.

your formula can be simplified to $$1/30\,n \left( 2\,n+1 \right)  \left( n+1 \right)  \left( 3\,{n}^{2}+3
\,n-1 \right) 
$$
for $n=1$ we have $$1^4=\frac{1}{30}\cdot 3\cdot 2\cdot 5$$ this is right
now we assume that $$1^4+2^4+3^4+...+k^4=\frac{1}{30}k(2k+1)(k+1)(3k^2+3k-1)$$
now we have to prove that
$$1+2^4+3^4+...+k^4+(k+1)^4=\frac{1}{30}(k+1)(2k+3)(k+2)(3(k+1)^2+3(k+1)-1)$$
have you got it?
now you have to Show that
$$(k+1)^4+\frac{1}{30}k(2k+1)(k+1)(3k^2+3k-1)=\frac{1}{30}(k+1)(2k+3)(k+2)(3(k+1)^2+3(k+1)-1)$$
can you finish?
