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.

Okay, I figured out where I was getting hung up. I was thinking $x$ would be any number and that $n$ meant infinity (I gotta stop doing that!?) 
So my thought process was like "Oh, if $n = 5$ then I have 5 solutions, but if $n = \infty$ then I have infinity solutions so that means there are infinity solutions for $x^n = 1$ this was super flawed :( Honestly, it seems like the question is a little wild.
So, for any kids who might stumble upon this question with the same ????? I had on my face - $n$ is just some number. Consider the easy fact that when $x \in \mathbb{R}, x^n = 1$ has one solution when $n$ is odd, and 2 solutions when $n$ is even (think about this!)
Add in the complex numbers and now you're looking at roots of unity which tells us.... what? ;) Check out those patterns too. if I choose $n = 5$ how many solutions are there for $x^5 = 1$ with $x \in \mathbb{C}$?
Everyone gave great hints, I think I just didn't really know what I needed to know until I knew it.
Feel good knowing that it (hopefully) didn't take you two days of back and forth's with a mathematician to figure it out.
