Converting $(1+...+n)^2*(n+1)^3$ to $(2+...+2n)^2$ I'm currently going through Calculus by Spivak by myself, and came across a proof by induction requiring to prove $1^3+...+n^3 = (1+...+n)^2$
Naturally, to prove this, I need to somehow convert $(1+...+n)^2+(n+1)^3$ to $(2+...+2n)^2$.
After quite a bit of thinking, I'm still not sure how to do this. I think i may be forgetting about some property of squares that we're supposed to be using.
Note: Please only provide a hint, not the complete answer. 
Edit: I was mistakenly taking $(1+...+n)^2*(n+1)^3$ rather than $(1+...+n)^2+(n+1)^3$. Corrected.


Naturally, to prove this, i need to somehow convert $(1+...+n)^2*(n+1)^3$ to $(2+...+2n)^2$.

Unfortunately, "$(1+...+n)^2*(n+1)^3 = (2+...+2n)^2$" is not true for all $n$. Once you simplify the summations, the left side will be a polynomial with degree $2 \cdot 2 + 3 = 7$ while the right side will be a polynomial with degree $2 \cdot 2 = 4$.
The induction step is to assume that $$1^3+2^3+\cdots+n^3 = (1+2+\cdots+n)^2$$
 for some integer $n$, and then prove that $$1^3+2^3+\cdots+n^3+(n+1)^3 = (1+2+\cdots+n+(n+1))^2.$$
As Arnie Dris suggested, the identity $1+2+\cdots+n = \dfrac{n(n+1)}{2}$ is useful. 
Note that \begin{align*}& (1+2+\cdots+(n+1))^2 - (1+2+\cdots+n)^2 \\ &= \left[\dfrac{(n+1)(n+2)}{2}\right]^2-\left[\dfrac{n(n+1)}{2}\right]^2 \\ &=\dfrac{(n+1)^2}{4}\left[(n+2)^2-n^2\right] \\ &= \dfrac{(n+1)^2}{4}[4n+4] \\ &= (n+1)^3\end{align*}
Do you see how to complete the proof using this?
