Finding Polynomial Equations for $1^4 + 2^4 + 3^4 + \ldots n^4$ Find a polynomial expression for
$$ 1^4 + 2^4 + 3^4 + ... + n^4 $$
I know you have to use the big theorem but I can't figure out how you would start to compute the differences. Suggestions?

Note that
\begin{aligned}
\sum_{i=1}^{n}i^5 &= \sum_{i=0}^{n-1}(i+1)^5 \\ 
& =  \sum_{i=0}^{n-1}\left(i^5 + 5i^4 + 10i^3 + 10i^2+5i+1 \right) \\
&= \sum_{i=0}^{n-1}i^5 + \sum_{i=0}^{n-1}\left(5i^4 + 10i^3 + 10i^2+5i+1 \right) \\
&= -n^5 + \sum_{\bbox[yellow]{i=1}}^{\bbox[yellow]{n}}i^5 + \sum_{i=0}^{n-1}\left(5i^4 + 10i^3 + 10i^2+5i+1 \right) \\
\end{aligned}
After canceling out the expression $\sum i^5$ from the sides, we'll get
$$n^5=\sum_{i=0}^{n-1}\left(5i^4 + 10i^3 + 10i^2+5i+1 \right)$$
Which can be rewritten as 
\begin{aligned}
5\sum_{i=0}^{n-1}i^4 &= n^5 - \sum_{i=0}^{n-1}\left(10i^3 + 10i^2+5i+1 \right)\\
\sum_{i=0}^{n-1}i^4 &=\frac15\left(n^5- 10\sum_{i=0}^{n-1}i^3 -10\sum_{i=0}^{n-1}i^2 -5\sum_{i=0}^{n-1}i -n \right) 
\end{aligned}
By defining $S_k=\sum_{i=0}^{n-1}{i^k}$, this is equivalent to
$$S_4 =\frac15\left(n^5- 10S_3 -10S_2 -5S_1 -n \right) $$
or in the form below to resemble the similarity with the binomial expansion ($S_0=n$)
$$ 5S_4 +10S_3+10S_2+5S_1 + S_0 =n^5 $$
So, to find $S_4$, you just need to replace the expressions for $S_3$, $S_2$ and $S_1$, or repeat the above process to manually derive them.
