solutions for $x^n = 1$ I'm supposed to solve this in terms of $n$, a natural number.
I'm really getting tripped up on this, and I don't really know why. The only way this can have a solution is if $n = 0$, specifically the algebra I wrote to show this is
$
\begin{gather*}
x^n = 1\\
\log_x(x^n) = \log_x(1)\\
n\log_x(x) = 0\\
n = 0
\end{gather*}
$
So, does this mean there are infinite solutions or just one unique solution? I'm not sure how this would change using $x$ as a complex number.

Your question is a bit unclear. You say that you are solving this "in terms of $n$, a natural number. The following doesn't give you all the details, you will need to figure out what the question is actually asking before you write down your solution. 
First question is: What is your definition of natural number? Or to put it differently, is $0$ a natural number. Let me assume that it is.
Second question is: Is $x$ a constant and you are solving for $n$ or is $n$ a constant and you are solving for $x$?
Case 1a: $x$ is positive constant not equal to $1$ and you are solving for $n$. In this case the only solution is $n=0$. For all other $n$, $x^n \neq 1$.
Case 1b: $x =1$ and you are solving for $n$. Here any $n$ will work.
Case 1c: $x$ is a negative constant and you are solving for $n$. In this case you have $x = -y$ for some positive $y$. So $x^n = (-1)^ny^n$. For this to be $1$ you need $(-1)^n = 1$ (otherwise the whole thing is negative because $y^n$ is always positive). This is satisfied exactly for the even numbers (including $0$). Then you need $y^n = 1$. If $y =1$, then this is true for all $n$ and your solution set is the set of all even $n$. If $y\neq 1$, then $n=0$ is the only solution.
Case 2a: $n$ is a constant (natural number greater than $0$) and you are solving for $x$. Let's assume that we are looking for solutions in the complex numbers. Here the solution set is know as the $n$th roots of unity. If
$$
x_0 = e^{2\pi i/n}
$$
The the so $n$ solutions are
$$
x_0, x_0^2, \dots, x_0^{n}.
$$
If you are just solving within the real numbers, then you only get those (from the list) that are real. This will be $-1$ and $1$. You only get $-1$ when $n$ is even.
Case 2b: $n=0$ and you are solving for $x$. In this case the solution set is all real (or complex) numbers except $0$. While some are going to disagree with me on this, $0^0$ is commonly not defined. (Just ask your teacher what convention you use in the class.)

Note here the case where you are solving for $n$ and $x\neq 1$ is a positive constant. In this case we got that $n=0$. And here your solution is actually correct! (within the real numbers.)
