The following isn't a rigorous proof, but I think it's "aesthetic", and "rise[s] naturally from the ground", as the original question asked for.

In searching for intuition for Taylor Series, I've developed a perspective involving Pascal's Triangle, which arises from recursively applied Riemann Sum approximations to the function.

I found @Bob Pego's answer really helpful and it's how I started developing this.

The end result involves coefficients based on rows of Pascal's Triangle, and the sequence of approximations (sequence of rows) looks like this

"Pascal" approximations for sin(x)

And they're much less efficient approximations than plain finite Taylor polynomial

Taylor approximations for sin(x)

I'll explain the derivation, but the essence of it is that the recursive Riemann Sum procedure produces binomial coefficients -- rows of Pascal's Triangle -- which are also simplex numbers. Simplex numbers converge to factorial fractions of hypercubes. The nth triangle number approaches $n^2 / (2!)$, the nth tetrahedral number approaches $n^3 / (3!)$, and so on.

A regular Riemann Sum approximation of f(x) of "resolution" 4 would be

$$
f(x) \approx f(0) + f'(0) \cdot \frac x4 + f'(x/4) \cdot \frac x4 + f'(2 \cdot x/4) \cdot \frac x4 + f'(3 \cdot x/4) \cdot \frac x4
\\ = f(0) + \frac x4 \cdot (f'(0) + f'(x/4) + f'(2 \cdot x/4) + f'(3 \cdot x/4))
$$

After each discrete step, we update the slope by setting it to the true slope of the function -- what the 1st derivative is *at* that point we've stepped to along x. This is the idea of a Riemann Sum.

But since we're interested in Taylor Series (about 0) here, let's pretend that we can't update to $f'(x/4)$ directly, and can only use the values of all derivatives evaluated at 0, not at $x/4$ or anywhere else.

So instead of updating to the actual slope, we'll use a recursive approximation to get an approximate slope update. We can now recurse and approximate each of the terms that have a non-0 x value. For example,

$$
f'(3x/4) \approx f'(0) + \frac x4 \cdot (f^{(2)}(0) + f^{(2)}(x/4) + f^{(2)}(2 \cdot x/4))
$$

There are still some terms with $f$ evaluated elsewhere than 0, so we recursively approximate terms until all terms are derivatives of $f$ evaluated at 0.

For resolution 4, you'll end up with

$$
f(x) \approx 1 \cdot (\frac x4)^0 \cdot f(0) + 4 \cdot (\frac x4)^1 \cdot f'(0) + 6 \cdot (\frac x4)^2 \cdot f^{(2)}(0) + 4 \cdot (\frac x4)^3 \cdot f^{(3)}(0) + 1 \cdot (\frac x4)^4 \cdot f^{(4)}(0)
$$

Note the appearance of the Pascal row $1, 4, 6, 4, 1$.

In general, for resolution n, that will be

$$
f(x) \approx \sum_{k=0}^n {n \choose k} \frac {x^k} {n^k} f^{(k)}(0)
$$

But I prefer to focus on the simplex perspective. Equivalently, that's

$$
f(x) \approx f(0) + \frac {natural_n}{n} f'(0) x + \frac {triang_{n - 1}}{n^2} f^{(2)}(0) x^2 + \frac {tetra_{n - 2}}{n^3} f^{(3)}(0) x^3 + \frac {penta_{n - 3}}{n^4} f^{(4)}(0) x^4 + ...
$$

Where e.g. $penta_{n - 3}$ is the $(n - 3)th$ pentatope number, like if we index the simplex numbers from 1 to infinity. A few examples:

$$
\color{blue}{triang}_{\color{red}{1}} = {(\color{blue}{2} - 1) + \color{red}{1} \choose \color{blue}{2}} = 1, \color{blue}{triang}_{\color{red}{4}} = {(\color{blue}{2} - 1) + \color{red}{4} \choose \color{blue}{2}} = 10
$$

$$
\color{blue}{tetra}_\color{red}{2} = {(\color{blue}{3} - 1) + \color{red}{2} \choose \color{blue}{3}} = 4, \color{blue}{tetra}_\color{red}{5} = {(\color{blue}{3} - 1) + \color{red}{5} \choose \color{blue}{3}} = 35
$$

etc.

Check the Pascal's Triangle wikipedia page if you're not following that.

**Simplex numbers approaching factorial fractions of hypercubes**

$$
\frac {tetra_{n - 2}} {n^3} = \frac {{(3 - 1) + (n - 2) \choose 3}} {n^3}
= \frac {n \choose 3} {n^3} = \frac {\frac {n (n - 1) (n - 2)} {3!}} {n^3}
= \frac {n (n - 1) (n - 2)} {n^3} \cdot \frac {1} {3!}
$$

and

$$
\lim_{n \to \infty} \frac {n (n - 1) (n - 2)} {n^3} \cdot \frac {1} {3!} = \frac {1} {3!}
$$

Taking $n$ to $\infty$ corresponds to increasing the "resolution" of your Riemann Sum, and approaching continuous integration, thus approaching the Taylor Series.

Just like this "triangle"

```
0
00
000
0000
```

Is a low resolution of an actual right isosceles triangle *polygon*

This may seem really roundabout given the concise alternative of the ${n \choose k}$ binomial coefficient notation, but I think simplexes are a nice way to visualize the "lagged" effect of higher order derivatives. If you begin traveling with constant acceleration of 1, then after 1 unit of time, your displacement will be the area of the right triangle in a unit square, $1/2! = 1/2$. If you begin traveling with a constant jerk of 1, then after 1 unit of time, your displacement will be the area of a tetrahedron in the corner of a unit cube, $1/3!$ = $1/6$.