Consider the following excerpt from Nets Katz's notes on calculus here, from pages 5 and 6.
As you can imagine, the argument generalizes to the sum of $k$th powers.$$S_k(n) = \sum_{j = 1}^n j^k.$$Keeping track of $k + 1$-tuples with some entries the same is tricky, but the highest order term in the formala is easy to guess. Gauss's trick is that $k + 1$-tuples with a single largest component have that component in one of $k + 1$ places. So what we get is that$$S_k(n) = {1\over{k + 1}}n^{k + 1} + \text{lower order terms.}$$You guys know some calculus so this should be familiar to you as one of the most basic facts in calculus. It encodes that the indefinite integral of $x^k$ is ${1\over{k + 1}}x^{k + 1}$. That's the same factor of ${1\over{k+1}}$ in both places. So what's actually happening is that Gauss' trick gives you a new (and perhaps unfamiliar) way of proving this fundamental fact. What is the way you're used to deriving it? Basically you work through the fundamental theorem of calculus, you know how to take derivatives and you know you have to guess a function whose derivative is $x^k$.
Question. How are these different proofs related, and if possible, is there a way to visualize the relation between these different proofs?
Only a comment, because I'm not sure that I understand what you like to know.
$\,(A)\enspace$ With the Bernoulli polynomials we have the following formula :
$$ S_k(n)=\frac{B_{k+1}(n+1)-B_{k+1}(1)}{k+1}$$
$B_k(x)\,$ is a polynomial of $\,x\,$ of degree $\,k$ .
We have $\enspace B_0(x)=1\,$ , $\enspace\displaystyle\frac{d}{dx}B_k(x)=k B_{k-1}(x)\enspace$ and $\enspace\displaystyle\int\limits_0^1 B_k(t)dt=0\enspace$ for $\,k\geq 1\,$ .
It follows $\,B_k(x+1)-B_k(x)=k x^{k-1}\,$ and therefore the summation formula.
$\,(B)\enspace$ If you are looking for a generalization of the summation by differences for $S_k(n)$,
$\hspace{1cm}$ you will find that in Concrete Mathematics.
For a polynom $\,f_m(x)\,$ of degree $\,m\,$ we have with the method of differences the summation formula $$\sum\limits_{j=0}^n f_m(j) = \sum\limits_{k=0}^m {\binom {n+1}{k+1}} \sum\limits_{v=0}^k (-1)^{k-v} {\binom k v} f_m(v) \enspace\enspace .$$
With $\,f_m(j):=j^m\,$ one gets the well-know formula for $\,S_m(n)\,$ with the Stirling numbers of the second kind.
Some hints:
Given: $\enspace\Delta f(x):=f(x+1)-f(x)\,$, $\enspace If(x):=f(x)\,$ , $\enspace Ef(x):=f(x+1)$
with their generalisations
$\hspace{1.45cm}\Delta^{n+1} f(x):= \Delta(\Delta^n f(x))$ , $I^{n+1} f(x):=I (I^n f(x))$ and $E^{n+1} f(x):=E (E^n f(x)) \,$ .
Be $\,m\in\mathbb{N}_0\,$ and $\,v\in\{0,1,2,…,m\}\,$ .
For $\, x^{\underline{n}}:= x (x-1) … (x-n+1) \,$ follows $\,\Delta^v x^{\underline{m}}=m^{\underline{v}}x^{\underline{m-v}}\,$ .
And we get, written in a symbolic way (using the linearity of the difference $\,\Delta:=E-I\,$ which
means $\,\Delta f(x)=(E-I)f(x)=Ef(x)-If(x))\,$ to simplify the method of differences:
$$ \Delta^k f_m(x) = (E-I)^k f_m(x) = \sum\limits_{v=0}^k (-1)^{k-v} {\binom k v} E^v f_m(x) = \sum\limits_{v=0}^k (-1)^{k-v} {\binom k v} f_m(x+v) $$
Be $\,\displaystyle f_m(x) := \sum\limits_{k=0}^m b_k x^{\underline{k}}\,$ .
It follows $\,\displaystyle \Delta^v f_m(x) = \sum\limits_{k=0}^m b_k \Delta^v x^{\underline{k}} = \sum\limits_{k=0}^m b_k k^{\underline{v}} x^{\underline{k-v}} \,$ and the special case $\,\Delta^k f_m(0)= b_k k! \,$ .
Substituting $\, b_k\,$ by $\, \displaystyle \frac{\Delta^k f_m(0)}{k!}\,$ and using the formula for $\,\Delta^k f_m(x)\,$ with $\,x:=0\,$ gives us the summation formula
$$f_m(x)=\sum\limits_{k=0}^m \frac{x^{\underline{k}}}{k!} \Delta^k f_m(0)= \sum\limits_{k=0}^m {\binom x k} \sum\limits_{v=0}^k (-1)^{k-v} {\binom k v} f_m(v) \enspace .$$
Finally $\,x\,$ substituted by the counter $\,j\in\{0,1,2,…,n\}\,$ with $\,n\in\mathbb{N}_0\,$ and
$\displaystyle \sum\limits_{j=0}^n {\binom j k} = {\binom {n+1}{k+1}}\,$ we get the formula for $\,\displaystyle\sum\limits_{j=0}^n f_m(j)\,$ above.
For further informations please see Concrete Mathematics .