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Picture proof of 

Consider the first n rows and columns of a square multiplication table; in our example we will let n=5.  What is the sum of all the numbers in this grid?

 1  2  3  4  5    = 1×15
 2  4  6  8 10    = 2×15
 3  6  9 12 15    = 3×15
 4  8 12 16 20    = 4×15
 5 10 15 20 25    = 5×15
                                    = 15×15

Of course, the sum of the first row has to be a triangular number, and when we square it when written as a binomial coefficient, we get the left side of the equation we are trying to prove.

Now we have to get the same sum to equal 1³ + 2³ + 3³ + 4³ + 5³; we will use color on the same matrix to show how we get the cubes.

 1  2  3  4  5   1 = 1³
 2  4  6  8 10   2+4+2 = 2³
 3  6  9 12 15   3+6+9+6+3 = 3³
 4  8 12 16 20   4+8+12+16+12+8+4 = 4³
 5 10 15 20 25   5+10+15+20+25+20+15+10+5 = 5³

We leave it as an exercise for the reader to prove why taking these 2n-1 entries from row n and column n will always add up to n³.  (Hint.)