A formula for $1^4+2^4+...+n^4$ I know that 
$$\sum^n_{i=1}i^2=\frac{1}{6}n(n+1)(2n+1)$$
and 
$$\sum^n_{i=1}i^3=\left(\sum^n_{i=1}i\right)^2.$$
Here is the question: is there a formula for 
$$\sum^n_{i=1}i^4.$$

yes there is a formula, $$\sum_{i=1}^ni^4=1/30\,n \left( 2\,n+1 \right)  \left( n+1 \right)  \left( 3\,{n}^{2}+3
\,n-1 \right) 
$$
