Validity of Proof of Sum of First $n$ Natural Numbers Background
Lately I've been self-studying Tom M. Apostol's Vol. 1 Calculus to make my understanding of the subject more rigorous after taking the actual class. I came across a proof for what the sum of the squares of the first $n$ natural numbers was - $$\sum_{i=1}^n i^2$$ and how it was equal to
$$\frac {n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\,.$$
In short, this involved the notion of a telescoping series, but I'd rather not go too deeply into the specifics. Rather, my interest turned to that of the sum of the first $n$ natural numbers, which was involved in the aforementioned demonstration.

Motivation
I'm sure plenty of you all have seen at least one proof of the following equality -
$$\sum_{i=1}^n i = 1 + 2 + \cdots + (n - 1) + n = \frac{n(n+1)}{2}\tag{1}\label{1}\\$$
Examples of these include a visual proof, proof by induction, etc. The purpose of this post is to explain my proof, whether it is valid, and how I could improve it if so. (Feel free to also critique my notation, I'd appreciate it.)
If done correctly, we should be able to find what the sum is equal to without making what I believe to be huge intuitive leaps.

Proof
Starting with \eqref{1}, we set the sum equal to a new variable $k$.
$$1 + 2 + \cdots + (n - 1) + n = k\tag{2}\label{2}\\$$
From here, one can realize that in some sense all the terms leading up to $n$ are fairly "close" in value to it; in other words, they can be put into the form $(n-a)$. As a result, \eqref{2} becomes -
$$(n - (n - 1)) + (n - (n - 2)) + \cdots + (n - 2) + (n - 1) + n$$
Rearranging the many $n$ together as well as the remaining terms, we get -
$$\underbrace{n + n + \cdots + n}_{n\text{ times}} + [-1 - 2 - \cdots - (n - 2) - (n -1)]$$
$n$ added unto itself $n$ times is the definition of $n^2$. This and factoring out the negative from the remaining sum gives -
$$n^2 - [1 + 2 + \cdots + (n - 2) + (n -1)]$$
Recall that this is still equal to the original sum, $k$. Notice as well that the remaining sum is equal to $k$ minus the $n$ term. Our equation transforms into -
$$n^2 - (k - n) = k \tag{3}\label{3}\\$$
Solving for $k$ (the original sum) results in -
$$k = \frac{n^2 + n}{2} = \frac{n(n + 1)}{2} \tag{4}\label{4}\\$$
Hence proving \eqref{1}.
Is this logically sound, or did I mess up somewhere?
 A: This is correct but can be made simpler:
$$S=1+2+3+\cdots(n-1)+n$$
and
$$S=n+(n-1)+(n-2)+\cdots2+1,$$
so that by addition
$$2S=(n+1)+(n+1)+(n+1)+\cdots(n+1)+(n+1)=n(n+1).$$
A: A very different method:
The general term is $i$, and we can write
$$S_i-S_{i-1}=i,$$ which is a linear polynomial in $i$. Now as the backward difference of a quadratic polynomial is a linear polynomial, we must have
$$S_i=ai^2+bi+c.$$
The coefficients can be found by identification, using the first values of $S$:
$$0^2a+0b+c=0,\\1^2a+1b+c=1,\\2^2a+2b+c=3,$$
giving
$$a=b=\dfrac12,c=0.$$
It is worth to note that we can easily generalize to the sum of $i^2$ and following.

$$0^3a+0^2b+0c+d=0,\\1^3a+1^2b+1c+d=1,\\2^3a+2^2b+2c+d=5,\\3^3a+3^2b+3c+d=14.$$$$a=\dfrac13,b=\dfrac12,c=\dfrac16,d=0.$$

A: This response may be somewhat long-winded, so I am formatting it as an answer.
I also own Apostol's "Calculus" and consider it a jewel.  However, the point of his method of attacking sums of powers early in his Volume 1 is simply to provide an example of what is possible.
This topic can be considered as an entry point into the sub-topic of "Bernoulli numbers".  This sub-topic has both a Calculus and a non-Calculus (i.e. algebra only) component.
I argue that Apostol's treatment of the algebra only portion of Bernoulli numbers is somewhat inferior.  In contrast, see
General form for sum of powers.
The above approach allows one to use algebra only (no Calculus) to directly attack (i.e. directly derive) a formula for  $\sum_{i=1}^n i^k$ for any $k \in \mathbb{Z^+}.$  In contrast, Apostol's approach is recursive.
Again, although Apostol's approach is useful as an example of the power of analysis, it is (perhaps) not the best approach to the algebra only portion of the sub-topic "Bernoulli numbers".
