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: 
I guess we agree on $
e^{a + i \phi} = e^{a} \cdot e^{i \phi}
$. So let's look at the latter term with imaginary exponents.

Imaginary exponents are indeed of a different quality then real ones. Maybe your best approach is to realize that the justification for the "magic" Euler equation

$$
e^{i \phi} = \cos(\phi) + i \sin(\phi)
$$
only comes from the formal continuation of the exponential function from real to complex domain. Indeed, by just taking the series expansion of the exponential function, $e^x = \sum\_{n=0}^\infty x^n/n!$, apply for $x = i \phi$, and grouping real and imaginary parts you see that this holds true. This gives you (for odd and even powers of $\phi$), with real $\phi$, of course: 

$$
e^{i \phi} = \sum\_{n=0}^\infty (i\phi)^n/n! = \sum\_{n=0}^\infty (-1)^n \phi^{2n}/(2n)! + i \sum\_{n=0}^\infty (-1)^n \phi^{2n +1}/(2n+1)!
$$
and the two sums are just the expansions of $\cos \phi$ and $\sin \phi$, for real $\phi$.

So yes, it is unintuitive that you now stay on the unit circle in the complex domain, that it's cyclic in $2 \pi$, etc.

---

EDIT: (see also Paul Sinclair's comment)

If you know about the derivatives of $\cos \phi$ and $\sin \phi$, here's an approach without series expansions.

We know that $\cos' \phi = - \sin \phi$, and $\sin' \phi =\cos \phi$. Now define a function
$
f(\phi) = \cos(\phi) + i \sin(\phi)
$.
Then you have $
f'(\phi) = - \sin(\phi) + i \cos(\phi) = i f(\phi)
$. 

On the other hand, we know for real constants $a$ that the equation $
f'(\phi) = a f(\phi)
$ has a unique solution, which is $f(\phi) = c \cdot e^{a \phi}$ with some constant $c$ which is determined by some initial condition. Now comes the trick again of the formal continuation of the exponential function from real to complex domain. We allow now that $a$ may as well be complex. Then we have that $
f'(\phi) = i f(\phi)
$ (as taken from the trig function above) has the unique solution $f(\phi) = c \cdot e^{i \phi}$. Determine the constant $c=1$ by observing $f(0) = c \cdot e^{i 0} = c$ and also $
f(0) = \cos(0) + i \sin(0) = 1
$. 

So this explains Euler's equation as well.
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.