Why can one take the power of $e$ directly? The definition of Euler's constant to the power $x$, $e^x$, is
$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + {...}$$
And of course, we have the number $e$ defined as
$$e = \sum_{n=0}^\infty \frac{1}{n!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + {...}$$
or
$$e = \lim_{n\to \infty} (1+\frac{1}{n})^{n}$$
$e$ and $e^x$ here are expressions of a sum of infinite series. When one calculate $e^x$, he doesn't go by the definition of $e^x$, but instead calculates the numerical value of $e$, and takes the power of that numerical value directly.
How can one simply take the power of the numerical value of $e$ directly, and be sure the answer is $e^x$? And what about in the context of arbitrary powers of $e$?
p.s There are also different definitions of $e$, like:
$$\int_1^{e^x}{\frac{1}{t}dt}=x$$
$$\frac{d}{dx}e^x = e^x$$
$$\frac{d}{dx}log_e{x}=\frac{1}{x}$$
But they do not explain the concern too.
 A: I assume that the question is (with the definition of the exponential function $\exp(x):=\sum\limits_{k=0}^{\infty} \frac{x^k}{k!}$):

Why do we have $\exp(1)^x = \exp(x)$?

There are (at least) two definitions of powers $a^x$ of real numbers. The first one uses $\exp$ and makes the claim trivial. The second (and probably more natural) one first defines $a^x$ when $x$ is an integer, then when $x$ is a rational number and finally when $x$ is a real number. I won't explain the details, because these are contained in every book on analysis.
So let us verify $\exp(1)^x = \exp(x)$ with this definition. We will only need the formula
$$\exp(x+y)=\exp(x) \cdot \exp(y).$$
It immediately implies by induction $\exp(1)^x = \exp(x)$ when $x$ is an integer. If $x=p/q$ is rational, it follows
$$(\exp(1)^x)^q = \exp(1)^p = \exp(p)=\exp(x)^q$$
and hence $\exp(1)^x = \exp(x)$. Finally, if $x$ is a real number, there is a sequence of rationals $x_i$ convering to $x$. Hence,
$$\exp(1)^x = \lim_i \, \exp(1)^{x_i} = \lim_i \, \exp(x_i) = \exp(x).$$
A: When you say calculate the value of $e$ and then take the power $e^x$, what does taking a power $a^x$ mean? By definition, we let
$$
a^x = \exp(x\log(a)),
$$
where $\exp$ is defined as the power series you mentioned and $\log$ is its left inverse. Thus,
$$
e^x = \exp(x\log(e)) = \exp(x\log(\exp(1)) = \exp(x),
$$
since $e$ is defined as $\exp(1)$.
A: Here is a proof of the equivalence of the two definitions. I take 
$$e=\lim\limits_{n \to\infty}  \left(1+\frac{1}{n}\right)^n.$$
as the definition of $e$ and denote by $e^x$ the ordinary exponent function ($a^b$ can be defined directly for all $a>0$ and $b$) now we show 
$$e^x=\lim\limits_{n \to\infty}  \left(1+\frac{x}{n}\right)^n.$$
This follows by writing
$$\left(1+\frac{x}{n}\right)^n= \left[ 
\left(1+\frac{1}{n/x}\right)^{n/x}\right]^x$$
And $$\left(1+\frac{1}{n/x}\right)^{n/x}\to e$$ so taking exponents we get the result.
