Conditions on exponential function with natural exponent in the case of series of functions To define a real exponential function
$$f(x)=a^x=e^{x \,\mathrm{lg} a}$$
It is strictly necessary that $a>0$.
But is the same true if the exponent is a natural number?

In function series and moreover in power series I see things like
$$\sum_{n\geq0} f(n)  x^n, \,\,\,\, x\in \mathbb{R}$$
So there is not the resctriction $x>0$.
It makes sense because natural number in the exponent do not create problems like even roots of negative numbers, so I think I'm ok with this.
Nevertheless if I rewrite $x^n$ using the definition of logaritm (for example this passage is useful while evaluating limits of such expressions), I get
$$\sum_{n\geq0} f(n) \,\, e^{n \, \mathrm{lg}x}$$
So suddenly I get a $\mathrm{lg}x$ and I need the condition $x>0$. 
On the one hand this passage should be valid because it is just the definition of logaritm, but on the other hand it looks like I cannot perform it without meet a restriction on $x$ that was not present in the form $x^n$.

So what is the point here? Is the use of the definition of logaritm not allowed in these cases, or is it wrong to avoid setting the condition $x>0$ in $x^n$?
 A: 
On the one hand this passage should be valid because it is just the definition of logaritm

No, it is not valid in general, since $x^n = e^{n \ln x}$ is only valid for $x \gt 0$.
One could technically use $x = |x| \cdot \operatorname{sgn}(x)$ to write it as $$x^n = e^{n \ln |x|} \cdot \operatorname{sgn}(x)^n \quad \forall \;x \in \mathbb{R} \setminus \{0\}, \;n \in \mathbb{Z} $$
though that would look quite odd in practice. Most often, one would use an appropriate substitution to first remap the variable to a positive domain, instead.
A: The key is to realize that there are two distinct exponentiation functions, where the one becomes a motivating model for (but does not necessarily embed in) the other.  These are unfortunately often conflated, and the identical notation does not help.
The first is the case of natural number exponents ($\Bbb N_+$ is the positive integers, and $X$ is any power-associative magma, which you can take as $\Bbb R$.  If the identity element $1$ is defined, it can be extended to $n=0$, and if $x$ has a multiplicative inverse, to all $n$.):
$$X \times \Bbb{N}_+ \to X : (x,n) \mapsto \prod_{i=1}^n x$$
The second is the case of a base that is in $\Bbb{R}_+$, the positive reals, but the exponent can be from any real algebra $A$:
$$\Bbb{R}_+ \times A \to A : (x, y) \mapsto \exp(y \ln x)$$
In both cases, you can derive single-parameter functions by fixing one parameter, but this does not change the conclusion. 
You will see that the added restriction arises when you try to use the second in place of the first, but the substitution is not valid where the domain of the second mapping does not contain that of the first.
So the answer to your first question is no, it need not be true of integer exponents, depending on which definition you choose to use for when the exponent is restricted to the natural numbers.  The answer to your second question is that inc general, in expressions such as power series, the first case is usually the intended definition, not the second.
A: Of course if you use $e^{x\ln a}$ instead of $a^x$ then $a>0$ because otherwise the exponent $\ln a$ is not real any more. 
(1) It's $e^{x\ln a}=a^x$ for $a>0$ ;
(2) it's only $a^x$ (and not $e^{x\ln a}$) valid for $a\in\mathbb{R}\setminus\{0\}$ with $x\in\mathbb{Z}$. 
$e^{x\ln a}$ and $a^x$ are two different kind of constructions for evaluations and the value ranges have to be considered.
