Simple proof by induction: $1^3 + 2^3 +3^3 +...+ n^3 = [ (n(n+1))/2 ]^2 $ I am rather illiterate when it comes to mathematics, I am afraid. In an effort to change that, I grabbed a copy of 'What is mathematics? : An elementary approach to ideas and methods' and have already encountered some difficulty. It seems silly to waste so much time trying to solve it myself, so I decided to ask for some help in solving and illuminating the taken steps, so that I can solve some more on my own. I shall review some of the basics while awaiting answers, and hope that my session tomorrow shall be more productive...
The problem is as follows:
Prove by induction that $1^3 + 2^3 +3^3 ... n^3 = [ (n(n+1))/2 ]^2 $.
As is, at the point I decided to seek help and look up material for review, I have taken the following steps:
solved for the base case: n=1, $1^3 = [2/2]^2 $
$$ 1 = 1^2, 1=1 $$
Then after proving the basis, I stated the assumption that:
$1^3 + 2^3 ... k^3 = [ (k(k+1)) / 2 ]^2$ is true.
Then I tried to solve for the next case, $(k+1)^3$:  
$$ 1^3 + 2^3 + 3^3 ... k^3 + (k+1)^3 = [ ( (k(k+1)) / 2 ) + (k+1) ]^2 $$
$$ [ (k(k+1)) / 2 ]^2  + (k+1)^3   =  [ (k(k+1) + 1(k+1) ) / 2 ]^2 $$
at this point after failing for a while and having spent quite some time looking up material, I sought help. I hope I'm going down the right path here, but I suspect I shall soon find out... 
Help is greatly appreciated. 

You're on the right track! The simplest way to show the last line is to foil out the terms on the left and right and note that they're equal (tedious, I know--but it'll get you what you need). If you haven't done so already, try proving these formulas with induction (for extra practice; they're a little easier):
$$\sum_{k=1}^n k = \frac{n(n+1)}{2}\quad\quad \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}.$$
EDIT: And by "last line," I mean the "goal":
$$\left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3 = \left(\frac{(k+1)(k+2)}{2}\right)^2.$$
