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?

Even though this might be too much for simple task you want (you can assume the polynomial of fifth degree and calculate coefficients) I think it is worth mentioning that there is a general formula for finding polynomial of sum of $p$-th powers $1^p+2^p+\dots+n^p$, and it is called Faulhaber's formula.
In short 
$$1^p+2^p+\dots+n^p = \frac{1}{p+1} \sum_{j=0}^{p}(-1)^j\binom{p+1}{j}B_j n^{p+1-j}$$
where $B_j$ are called Bernoulli numbers, which you can compute for example as
$$
B_n = \sum_{k=0}^{n}\frac{1}{k+1}\sum_{r=0}^{k}(-1)^r \binom{k}{r} r^n
$$
giving first couple values $1,-1/2$,$1/6$,$0$,$-1/30$,$\dots$.
So in your case this becomes
\begin{align}
1^4+2^4+\dots+n^4 &= \frac{1}{5} \sum_{j=0}^{4}(-1)^j\binom{5}{j}B_j n^{5-j}\\
&= \frac{1}{5} ((-1)^0\binom{5}{0}B_j n^{5}+(-1)^1\binom{5}{1}B_1 n^{5-1}\\
&+(-1)^2\binom{5}{2}B_2 n^{5-2}+(-1)^3\binom{5}{3}B_3 n^{5-3}+(-1)^4\binom{5}{4}B_4 n^{5-4})\\
&= \frac{1}{5} (B_0 n^{5}-5 B_1 n^{4}+10 B_2 n^{3}-10 B_3 n^{2}+5 B_4 n)\\
&= \frac{1}{5} (n^{5}+\frac{5}{2}n^{4}+\frac{10}{6} n^{3}-\frac{5}{30} n)\\
\end{align}
