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. 

**HINT:**$$\left(\frac{k(k+1)}{2}\right)^2+(k+1)^3=\frac{k^2(k+1)^2}{4}+(k+1)^3$$$$=\frac{k^2(k+1)^2+4(k+1)^3}{4}$$$$=\frac{(k+1)^2(k^2+4(k+1))}{4}$$$$=\frac{(k+1)^2(k^2+4k+4)}{4}$$$$=\frac{(k+1)^2(k+2)^2}{4}$$Hopefully you can complete the proof from here...
