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
  


One technique is the Newton series, which permits termwise summation just as the Taylor series permits termwise integration. There is also the more powerful but less efficient method of indefinite summation, which is unsurprisingly the discrete version of integration by parts.
$
\def\zz{\mathbb{Z}}
\def\lbinom#1#2{{\large\binom{#1}{#2}}}
$

Let $R$ be the right-shift operator defined as:

$R = ( \text{function $f$ on $\zz$} \mapsto ( \text{int $n$} \mapsto f(n+1) ) )$

Namely, for any function $f$ on $\zz$ and $n \in Z$, we have $R(f)(n) = f(n+1)$.
For convenience we also define the summation operator:

$Σ = ( \text{function $f$ on $\zz$} \mapsto ( \text{int $n$} \mapsto \sum_{k=0}^{n-1} f(k) ) )$

Then we have the important property that $ΔΣ(f) = f$ for any function $f$ on $\zz$, analogous to the fundamental theorem of calculus.
Now we can derive summation by parts from the product rule in the same manner as for the continuous version. Consider any functions $f,g$ on $\zz$:

$Δ(f·g)(n) = f(n+1) g(n+1) - f(n) g(n) = f(n+1) Δ(g)(n) - Δ(f)(n) g(n)$ for any $n \in \zz$.

And hence the discrete product rule (with the usual pointwise sum and product of functions):

$Δ(f·g) = R(f)·Δ(g) + Δ(f)·g$.

Now by substituting $f$ with $Σ(f)$ and taking summation on both sides we get summation by parts:

$Δ(Σ(f)·g) = R(Σ(f))·Δ(g) + f·g$.
$Σ(f·g) = Σ(f)·g - Σ(R(Σ(f))·Δ(g)) + c$ for some constant function $c$ on $\zz$.


For convenience we let "$Δ_n E(n)$" denote "$E(n+1)-E(n)$" and "$Σ_n E(n)$" denote "$\sum_{k=0}^{n-1} E(k)$".
We can therefore compute $\sum_{k=0}^{n-1} k^3$ as follows:

$\sum_{k=0}^{n-1} k^3 = Σ_n\Big(\lbinom{n}{0}·n^3\Big)$   [Recall this definition of binomial coefficients.]
$\ = \lbinom{n}{1}·n^3 - Σ_n\Big(\lbinom{n+1}{1}·(3n^2+3n+1)\Big) + c$, where $c$ is some constant (independent of $n$),
$\ = \lbinom{n}{1}·n^3 - \lbinom{n+1}{2}·(3n^2+3n+1) + Σ_n\Big(\lbinom{n+2}{2}·(6n+6)\Big) + c'$,
    where $c'$ is some (other) constant,
$\ = \lbinom{n}{1}·n^3 - \lbinom{n+1}{2}·(3n^2+3n+1) + \lbinom{n+2}{3}·(6n+6) - Σ_n\Big(\lbinom{n+3}{3}·6\Big) + c''$,
    where $c''$ is some (other) constant,
$\ = \lbinom{n}{1}·n^3 - \lbinom{n+1}{2}·(3n^2+3n+1) + \lbinom{n+2}{3}·(6n+6) - \lbinom{n+3}{4}·6 + c''$.

By substituting $n = 0$ we readily get $c'' = 0$ and hence simplifying gives the desired formula.

Using this we can also easily compute things like $\sum_{k=0}^{n-1} k^3 3^k$ by applying it three times, each time reducing the degree of the polynomial part. Incidentally, there is another purely discrete technique to compute this sum, namely subtracting it from $3 \sum_{k=1}^n k^3 3^k = n^3 3^{n+1} + \sum_{k=1}^n (k-1)^3 3^k$, to reduce the polynomial degree, and repeating three times. But interestingly this technique does not work for $\sum_{k=1}^n k^3$.
