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 $\exp(a)\exp(b) = \exp(a + b)$.
Actually, I think your fundamental mistake is to think too fundamentally without thinking more fundamentally :P. The problem is deep rooted in how we define exponentiation. 

This used to confuse me for some time :) I hope this answer helps you and anybody else that might stumble upon my answer!

A Simpler Problem
=================

You seem comfortable calculating the value of $x^{1 \over b}$, but given a paper, a pen and infinite time, how would you evaluate:

$$3^{0.3}$$

How do you produce the digits? Most calculators reduce this expression to:

$$ e ^ {\log 3^{0.3}} = e ^ {0.3 \log 3}$$

And then we can use lookup tables to calculate $log3$ and indeed the exponential of the resulting number. The lookup tables are defined from limits and calculus though, so :

$$ e^x = \lim\_{n\rightarrow \infty} \left( 1 + {x\over n} \right)^n $$

and:

$$\log x = \int\_1^x {1 \over x} dx $$

So we can go ahead and work out this integral directly using a lookup as I mentioned, or by using something like Simpsons rule. 

Notice that we've lost the original definition here, we are no longer defining:

$$ a^b = \underbrace{a \* a \* a \* \ldots }\_\text{n times} $$

but instead, we are now defining exponentiation by:

$$ a^b = e ^ {b \log a }$$

This represents a fundamental shift in how we should perceive exponentiation. Because the idea of multiplying something by itself 0.3 times is not well defined, and we shouldn't think of it in that way anymore :)

Extending to Complex Numbers
============================

Now if we apply this definition to complex numbers, with the aide of Eulers amazing equation, that shows us multiplication of two complex numbers represents a multiplication of magnitudes and a rotation by their arguments, we can see that this equation is exactly what we would expect it to be:

$$ x ^ {i {\pi \over 4}} = e ^ {i \left( {\pi \over 4 }\log x \right)} = \cos\left( {\pi \over 4 }\log x \right) + i \sin \left( {\pi \over 4 }\log x \right)$$

This just uses our new definition of exponentiation, and it extends very naturally to complex numbers as you can hopefully see :)

Some Words
==========

I think some of the others put this better than I could. As you learn more and more mathematics, you start to learn to explain things that you once could only explain with a certain pattern of thinking, in terms of new patterns that would have been absurd to show a new learner.

I indeed asked myself this same question some time ago, and after some careful thought, I ended up with this :)

Ps.
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So I guess now you have to figure out how to calculate sin and cos ;) 

Have a good one!