Summation Identity: $\sum_{i=1}^ni^3 = \left( \frac{n(n+1)}{2} \right)^2$ I have to prove:
$$\sum\limits_{i = 1}^n i^3 = \Bigg( \frac{n(n+1)}{2}\Bigg)^2$$
Using the following:
$$n^3 = 6 {n \choose 3} + 6 {n \choose 2} + n \quad \forall n \in \mathbb{N}$$
My work is that first I substitute $n^3$ for $6 {n \choose 3} + 6 {n \choose 2} + n$. Then I go and invoke the sum over that (I am assuming this is how it works). That is,
$$\sum\limits^n_{i=1} \bigg(6 {i \choose 3} + 6 {i \choose 2} + i \bigg)$$
$$6 \sum\limits^n_{i=1} {i \choose 3} + 6 \sum\limits^n_{i=1} {i \choose 2} + \sum\limits^n_{i=1} {i \choose 1}$$
The summation identity is:
$$\sum\limits^n_{i=0} {i \choose k} = {n+1 \choose k+1}$$
invoking it over the sums yields:
$$6{n+1 \choose 4} + 6{n+1 \choose 3} + {n+1 \choose 2}$$
I think these are the right steps, but simplification seems a bit difficult to get the ending result.

Don't stop, just do! =)
$$
\begin{eqnarray*}
&& 6 \binom{n+1}{4} + 6\binom{n+1}{3} + \binom{n+1}{2} 
\\ &=& 6 \frac{(n+1) n (n-1)(n-2)}{24} + 6 \frac{(n+1) n (n-1)}{6} + \frac{(n+1) n}{2}
\\ &=& \frac{n(n+1)}{4} \left( (n-1)(n-2) + 4 (n-1) + 2 \right)
\\ &=& \cdots
\end{eqnarray*}
$$
I presume you can simplify the expression further to get the right hand side. 
