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 think it would be helpful for you to let go of the idea (at least in complex analysis) that $a^b$ means multiplying $a$ $b$ times by itself. Rather, exponentiation starts at the function $\exp : \mathbb C \to \mathbb C$, which you probably usually denote as $\exp(a) = e^a$. Then (once you have chosen a branch of the complex logarithm) you can define $a^b = \exp(\log(a) \cdot b)$. With that definition, all the repeated multiplication rules that you know come down to the identities $\exp(a + b) = \exp(a)\exp(b)$ and $\exp(0) = 1$. 

Rather than trying to press complex exponentiation into the mold of repeated multiplication, see complex exponentiation -- or more fundamentally, the function $\exp$ -- as its own thing, that *in special cases* can be interpreted as repeated multiplication, thanks to the functional relation $\exp(a)\exp(b) = \exp(a + b)$.
> 
> $x^n$ 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.