For your first question, ask yourself when x^e is undefined. You'll find it's defined everywhere. Your second question is a little deeper. I would read up on what exactly it means to have an irrational exponent.

To clarify, is $x$ a real or complex variable?

@whpowell96 my mistake, this question is considering the real numbers in the first part, and the complex in the second.

You should edit your question to reflect this since it is unclear as written

@whpowell96done!

So on my own, I went and did some sketching and think I answered the first part: $e = \sum_{n=0}^{\infty}\frac{1}{n!} \implies x^e = x^{2+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+...} = x^2 * x^{\frac{1}{2!}} * x^{\frac{1}{3!}}* x^{\frac{1}{4!}}*...$ and it is rather straightforward to show that this representation of the function has a nonnegative domain

You need to provide context for this question. Not just to satisfy the MSE Guidelines tour but also so that this will be more helpful for everyone. Where did it come from? In particular, what course is it, and have you learned about exponentiation? In particular, can you define $x^a$; $a$ a real number, for even positive $x$? [Assume $x^n$ and $x^{1/n}$ and $x^{-1}$ are defined for any integer $n$, and $x^ax^b=x^{a+b}$ and $(x^a )^m = x^{am}$.] If you cannot do so then IMO you have no business even considering this exercise.

@Mike It's a question I thought of spur of moment, it's not from any particular course, and yes I have learned of exponentiation. I find your comment to be astonishingly unhelpful and almost rude for someone with your spectacular comment history.

For $x<0$, the expression $x^e$ entails raising a negative number to a non integer exponent which leads the number to be complex. Therefore, $\text{dom}(f) = [0,\infty)$ for real numbers.

See this related post about the problem with negative numbers math.stackexchange.com/a/3356598/399263, completing the comment by CroCo just above.

@CroCo But isn't the domain of $f(x) = x^{\frac{1}{3}}$ all real numbers? at the very least, f(-27) = -3, and $\frac{1}{3}$ isn't an integer

@GMeso $e$ is not a rational with odd denominator, that is why.

@zwim, I read through the comments on that post and I found it very helpful, thank you!

$\sqrt[3]{-27} = -3$ because $(-3)^3= (-3)(-3)(-3) = -27$. Precalculus books cover this topic in detail. Students often fall into this trap.

@CroCo Yes, however it is still an act of raising a negative number to a non-integer power, and it is still defined, so it can't just be as simple as dom$(f)$ =[$0,\infty$) for $f(x) = x^a$ whenever $a \notin \mathbb{Z}$.

I would say "which leads the number in general to be complex". Though you got the point, @zwim clarified it further.

@CroCo, There are literally infinite rational numbers for which the domain is defined on the negative reals in this case. I get that it becomes a much messier topic once you include irrational and transcendental numbers, and with even denominator rationals you have the simple proof case for non negative domains, but your original comment said, quote, "raising a negative number to a non integer exponent (which) leads the number to be complex. Therefore, dom$(f)=[0,\infty)$ for real numbers" which logically implies that the domain of the cube root is only positive numbers, which is wrong.

@zwim: $e$ is not a rational with odd denominator, that is why --- Something that muddies the water even more is that "rational with an odd denominator" can be both true and not true for the SAME NUMBER. For example, $\frac{1}{3}$ and $\frac{2}{6}$ each denote the same number, but one is rational with an odd denominator and the other is rational with an even denominator. Thus, the usual precalculus interpretation of $(-27)^a$ for $a = \frac{1}{3}$ involves MORE than just the value of $a,$ it also involves the use of a specific representation of the number $a.$

Yes the fraction is supposed to be irreductible, i.e. $\frac pq$ with $\gcd(p,q)=1$ and $q$ odd. I detailled all that in my $(-1)^{2.16}$ post.