Compilation of proofs for the summation of natural squares and cubes I want to know different proofs for the following formulas,
$$
\sum_{i=1}^n{i^2} = \frac{(n)(n+1)(2n+1)}{6}
$$
$$
\sum_{i=1}^n{i^3} = \frac{n^2(n+1)^2}{2^2}
$$
Please do not mark this as duplicate, since what I specifically want is to be exposed to a variety of proofs using different techniques (I did not find such a compilation anywhere on the net)
I am only familiar with two proofs, one which uses expansion of $(x+1)^2 - x^2$ and $(x+1)^3 - x^3$ and the other which uses induction. I have provided link for the induction proof in a self-answer.
I am particularly interested in proof without words and proofs which use a unrelated mathematical concept (higher level math upto class 12 level is acceptable). 
Also, 
(1) Don't think I am being rude or anything, it is out of genuine interest that I am asking this question.
(2) Someone marked this as a duplicate of Methods to compute $\sum_{k=1}^nk^p$ without Faulhaber's formula
My question is different in three ways:
(i) I want to focus only on these two summation and not the general case,
(ii) Hence, it follows that the proofs which I am looking for a simpler than the ones provided in that link and are simpler (using images, pictures or high school algebra). What I want is to study new proofs. I believe it is a good practice when learning math to so this.
(iii) Since the proofs in the link are given for the general case, they are complicated and I am finding it hard to understand them. If someone is able to use the same method to the two cases in my question, then it would probably become much simpler and easier to digest.
Appendix
Feel free to make use of these topics in your answers, 
Calculus
Basic Binomial Expansion
Coordinate Geometry
Algebra (upto what 18 year olds learn)
Taylor Series Expansions
Geometry (18 year old level)
basically.............math which 18 year old's learn on Earth. 
If you want to err, then err on the higher math side:)

Answers Compilation List
  
  
*
  
*By Newton series
  
*By Sterling Numbers
  
*By Induction
  
*From the book Generatingfunctionology by Herbert Wilf
  
*By generalizing the following pattern
  $$\begin{align}
&\ \,4\cdot5\cdot6\cdot7\\=&(1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5+3\cdot4\cdot5\cdot6+4\cdot5\cdot6\cdot7)\\-&(0\cdot1\cdot2\cdot3+1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5+3\cdot4\cdot5\cdot6)\\
=&(1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+3\cdot4\cdot5\cdot4+4\cdot5\cdot6\cdot4)\\
\end{align}$$
  
*By Lagrangian Interpolation
  
*By Formal Differentiation
  
*By the Euler-Maclaurin Summation Formula
  
*By Assuming that the expression is a polynomial of degree $2$.
  
*A Proof Without Words for the cube case
  
*By integrating and assuming a error term.
  
*SimplyBeatifulArt's Personal Approach
  


By Lagrangian interpolation.
The expression of the sum of the cubes must be a quartic polynomial, because  its first order difference is cubic. Furthermore, it has no constant term because the sum of no numbers is zero.
Hence, the average of the cubes is a cubic polynomial, by the four points
$$(1,1),\left(2,\frac{1+2^3}2\right),\left(3,\frac{1+2^3+3^3}3\right),\left(4,\frac{1+2^3+3^3+4^4}4\right).$$
The resquested polynomial is the Lagrangian interpolant.

Average of ones (constant, by $(1,1)$):
$$\overline S_0(n)=1$$
Average of naturals (linear, by $(1,1)$ and $(2,\frac32)$):
$$\overline S_1(n)=\frac{\frac32-1}{2-1}(n-1)+1=\frac{n+1}2$$
Average of squares (quadratic, by $(1,1)$, $(2,\frac52)$ and $(3,\frac{14}3)$):
$$\overline S_2(n)=\cdots=\frac{2n^2+3n+1}2$$
Average of cubes (cubic, by $(1,1)$, $(2,\frac92)$, $(3,\frac{26}3)$ and $(4,\frac{45}2)$):
$$\overline S_3(n)=\cdots=\frac{n^3+2n^2+n}4$$

An explicit formula can be obtained for any degree, by means of the Lagrangian interpolation formula for equidistant points.
Newtonian interpolation could be easier.
