Sum of cubes of the first n natural numbers - Formula derivation

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Hello, Everyone. In this, we are going to derive the formula to find the sum of cubes of the first n natural numbers.

Let S = 13 + 23 + 33 + .......... + (n-1)3 + n3
(m+1)4 - m4 = (m4 + 4m3 + 6m2 + 4m + 1) - m4 = 4m3 + 6m2 + 4m + 1 ------------ (i)
put m = 1,2,3, ....... , (n-1), n in (i)
Then,
24 - 14  = 4(1)3 + 6(1)2 + 4(1) + 1 ------------ (1)
34 - 24  = 4(2)3 + 6(2)2 + 4(2) + 1 ------------ (2)
44 - 34  = 4(3)3 + 6(3)2 + 4(3) + 1 ------------ (3)
...................
n4 - (n-1)4  = 4(n-1)3 + 6(n-1)2 + 4(n-1) + 1 ------------ (n-1)
(n+1)4 - n4  = 4(n)3 + 6(n)2 + 4(n) + 1 ------------ (n)

Now, by adding equations (1),(2),(3), ...... ,(n-1) and (n), we get the following equation.
(n+1)4 - 14  = 4(13 + 23 + 33 + .......... + (n-1)3 + n3) + 6(12 + 22 + 32 + .......... + (n-1)2 + n2) + 4(1 + 2 + 3 + ......... + (n-1) + n) + (1 + 1 + 1 +........ n times)

1 + 2 + 3 + .......... + (n-1) + n = (n(n+1))/2 [Since sum of first n natural numbers = (n(n+1))/2] 
12 + 22 + 32 + .......... + (n-1)2 + n2 = [n(n + 1)(2n + 1)] / 6 [Since sum of squares of the first n natural numbers = [n(n + 1)(2n + 1)] / 6] and 
(n+1)4 - 14 = (n4 + 4n3 + 6n2 + 4n + 1) - 14 = n4 + 4n3 + 6n2 + 4n 

Therefore, (n+1)4 - 14 = n4 + 4n3 + 6n2 + 4n = 4S + 6((n(n + 1)(2n + 1))/ 6) + 4((n(n+1))/2) + n
n4 + 4n3 + 6n2 + 4n =  4S + 2n3 + 5n2 + 4n 
n4 + 2n3 + n2 = 4S
n2(n2 + 2n + 1) = 4S
n2(n2 + n + n +1) = 4S
n2(n(n + 1) + 1(n + 1)) = 4S
n2(n + 1)2 = 4S

Therefore, S = Sum of cubes of the first n natural numbers = (n(n + 1) / 2)2 
Therefore, formula to find the sum of cubes of the first n natural numbers = (n(n + 1) / 2)2 

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