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.
> 
> $x^n$ is just $n$ multiplications of $x$
> 
> 
> 

This is not *understanding exponentiation*; it is merely the reduction of exponentiation into other terms.

Reducing new concepts (in special cases) to familiar terms is a common *first step* to understanding something, but one eventually needs to proceed from there to synthesize an understanding about the concept itself, rather than use the other terms as a crutch.

I don't think you can *intuitively* understand complex exponentiation as a generalization of real exponentiation without first synthesizing an understanding of exponentiation.

You can, however, reduce complex exponentiation to other terms; the usual definition is

$$ z^w = \exp(w \log z)$$

so if you want to work with complex exponentiation in *other terms*, you can reduce the problem to understanding the complex exponential and logarithm.