Question: > 
> Claim:
>  $$\sum\_{k=0}^{n} \frac{x^k}{k!} < e^x, \qquad \forall x > 0$$
> 
> 
> 

Proof: 

Let $f(x)=e^x$. Then the Taylor series gives

$$f(x) = \sum\_{k=0}^{\infty} \frac{x^k}{k!}$$

$$= \sum\_{k=0}^{n} \frac{x^k}{k!} + \sum\_{k=n+1}^{\infty} \frac{x^k}{k!}$$

Now, because $x>0$, $$\sum\_{n+1}^{\infty} \frac{x^k}{k!} > 0$$

$$\implies f(x) > f(x) - \sum\_{n+1}^{\infty} \frac{x^k}{k!}$$

$$\implies \sum\_{k=0}^{n} \frac{x^k}{k!} = f(x) - \sum\_{n+1}^{\infty} \frac{x^k}{k!} < f(x)$$

QED.

I would appreciate constructive criticism on this proof.

Answer: 
My intuition about this works best when I consider it as manipulating the vector $(1,0)$ on the complex plane by performing two separate actions on it:

1. Scaling it based on the real part (with $\phi=0$ being the identity, positive values stretching it and negative values shrinking it).
2. Rotating it based on the imaginary part (this is the cool part).

So why does $e^{i \phi}$ perform a rotation? Well this comes out as a natural consequence of Euler's formula.

The simplest non-trivial example is of $\phi=\pi$, which gives us a rotation of $\pi$ radians, bringing the vector to $(-1,0)$ (this is a restatement of Euler's identity).

And in the general case, for any (real) value of $\phi$, Euler's formula gives us $e^{i \phi} = \cos\phi + i \sin\phi$, which has to be on the unit circle, as it's the vector $(\cos\phi, \sin\phi)$, the magnitude of which is:

$\|(\cos\phi, \sin\phi)\| = cos^2\phi + sin^2\phi = 1$

And this also helps us see that the effect of the imaginary part of the exponent is periodic, so if we want to, we can limit ourselves to only consider the range $\phi\in[0, 2\pi)$.

So to sum up, by performing a complex exponentiation, which is the combination of the scaling and rotation, we can get our vector anywhere on the complex plane, except for the origin (which is the limit when the real part tends to $-\infty$).

**Edit:**
Btw, if (Like me) you like infinities, you might find the concept of the [Riemann Sphere](https://en.wikipedia.org/wiki/Riemann_sphere) cool. The idea here is to add another "point at infinity" to the complex plane, and wrap the whole plane up as a sphere, with the point at infinity being exactly opposite from the origin. So when considering complex exponentiation on the sphere, scaling acts like changing the latitude, while rotation acts as changing the longitude. And scaling by $\pm\infty$ become well-defined, such that scaling by $-\infty$ takes the vector to the point at zero, while scaling by $+\infty$ takes it to the point at infinity, on the other side of the sphere. And in both these extremes, the rotation has no effect, which is very similar to how when navigating on Earth, when you're at the poles (latitudes $180^\circ N$ and $180^\circ S$), the longitude loses meaning.
A famous result in mathematics is Euler's Formula: $$e^{i\theta}=i\sin(\theta)+\cos(\theta)$$ Complex exponentials are used to parametize points in the complex plane so a point $$x+iy=re^{i\theta}=r(\cos(\theta)+i\sin(\theta))$$ So the function $e^{i\theta}$ can be viewed as a circle in the complex plane with radius $1$.

So complex exponentiation cannot be viewed as repeated multiplication anymore, and instead of stretching a point around the unit circle.

The reason e is the unique base that satisfies euler's formula has to do with the fact that the derivatives of $e^x$ and $cos(x),sin(x)$ cycle through the same values at a point.