Sum from $k=1$ to $n$ of $k^3$ $$\sum_{k=1}^n k^3 = \left(\frac{1}{2}n(n+1) \right)^2$$
I want to prove this using induction. I start with $(\frac{n}{2}(n+1))^2 + (n+1)^3$ and rewrite $(n+1)^3$ as $(n+1)(n+1)^2$, then factor out an $(n+1)^2$ from the expression:
$(n+1)^2((\frac{n}{2})^2 + (n+1))$
I'm confused where to go from here though.

$(n+1)^2((\frac{n}{2})^2 + (n+1))=(n+1)^2\frac{n^2+4n+4}{4}=(n+1)^2\frac{(n+2)^2}{2^2}= \left(\frac{1}{2}(n+1)((n+1)+1) \right)^2$
