Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction I recently proved that
$$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$
using mathematical induction.  I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs.

We begin by writing $k^3$ in a more clever fashion: $k^3 = k(k-1)(k-2) + 3k^2 - 2k$ :    
$$\sum_{k=0}^n k^3 = \sum_{k=0}^n k(k-1)(k-2) + 3k^2 -2k$$
Distributing the summation and adjusting our indices we obtain: $$ \sum_{k=3}^n k(k-1)(k-2) + \sum_{k=0}^n 3k^2 - \sum_{k=0}^n 2k$$ 
Notice, $$k(k-1)(k-2) = \frac {k!}{(k-3)!}$$
Now we have 
$$\sum_{k=3}^n \frac {k!}{(k-3)!} + \sum_{k=0}^n 3k^2 - \sum_{k=0}^n 2k$$
Notice that we have nearly obtained the binomial expansion of K choose 3, all we need to do is divide by 3! So we offset this by also taking the product of 3!
$$\sum_{k=3}^n \frac {k!}{(k-3)!} = 3!\sum_{k=3}^n \frac {k!}{(k-3)!3!} = 3!\sum_{k=3}^n\binom{k}{3} = 3!\binom{n+1}{4}$$
We have now obtained 
$$\sum_{k=0}^n k^3 = 3!\binom{n+1}{4} + 3\sum_{k=0}^n k^2 - 2\sum_{k=0}^n k$$
Focusing solely on the right-hand side we have
$$6\biggl(\frac {(n+1)!}{(n-3)!4!}\biggr) + 3\sum_{k=0}^n k^2 - 2\sum_{k=0}^n k$$ 
Assuming we already know the sum of the sequence of integers and squared integers (the 2 sums we have left) we have 
$$ \frac {(n+1)(n)(n-1)(n-2)}{4} + 3\frac{n(n+1)(2n+1)}{6} - 2\frac {n(n+1)}{2}$$
Generating common denominators and with a bit of algebra we now have
$$ \frac {n^4-2n^3-n^2+2n+4n^3+6n^2+2n-4n^2-4n}{4}$$
Combining like-terms we have reached our solution:
$$ \frac {n^4+2n^3+n^2}{4} = \biggl(\sum_{k=0}^n k\biggr)^2$$
