Expressing sums of powers I am trying to solve this question but I'm not really understanding how to continue, I would greatly appreciate some kind of tip.  
The Question
The formula for $1^2 + \cdots + n^2$ may be derived as follows.  We begin with the formula $$(k+1)^3 - k^3 = 3k^2 + 3k + 1$$
Writing this formula for $k=1,\ldots,n$ and adding we obtain an expression.
Thus we can find $\sum_{k=1}^{n}k^2$ if we already know $\sum_{k=1}^n k$.
Use this method to find $1^3 + \cdots + n^3$.
I am confused about this method.  How did they get the formula to begin with, and what formula am I supposed to use for $1^3 + \cdots + n^3$?
EDIT: I noted that they did already hint at $k^2$ being expressed from $k^1$, but we can't square $k^1$ into $k^3$

The formula is the Binomial Theorem, or just multiplication: $(k+1)^3=k^3+3k^2+3k+1$. This can be rewritten as 
$$(k+1)^3-k^3=3k^2+3k+1.$$
The corresponding formula for fourth powers is $(k+1)^4=k^4+4k^3+6k^2+4k+1$. It yields the identity 
$$(k+1)^4-k^4=4k^3+6k^2+4k+1.$$
That can be used to give a telescoping argument for the sum of the first $n$ cubes, much like the telescoping argument for the sum of the first $n$ squares. Sum both sides from $k=1$ to $k=n$. On the right, there is almost total cancellation, and we get
$$(n+1)^4-1=4\sum_1^n k^3+6\sum_1^n k^2+4\sum_1^n k +\sum_1^n 1.\tag{1}$$
Now a fair bit of messy algebra gets us $\sum_1^n k^3$, since we have formulas for every other sum in (1).  
In general, the Binomial Theorem is the assertion that if $m$ is a positive integer, then
$$\small (x+y)^m=\binom{m}{0}x^m +\binom{m}{1}x^{m-1}y+\binom{m}{2}x^{m-2}y^2+\cdots +\binom{m}{m-1}xy^{m-1}+\binom{m}{m}y^m.$$
Remark: It will turn out that 
$$1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2.$$
Once we have guessed this formula, it can be proved more easily by induction. 
