Why are there only 2 solutions for $x^n=1$? (where $n>0$)
I have been taught that an equation with the highest power $n$ will always have $n$ solutions. This does not appear to be the case with:
$$x^n=1 \implies x=\pm1$$
Where $n$ is even, the solutions are $1$ and $-1$; where $n$ is odd, the only solution is $1$.
Why? Is it a special case, or is there something I'm missing?
 A: It has $n$ roots, but only if you consider roots $x_i\in \Bbb C$. These "special" ones are calleds the roots of unity.
As an example, consider $X^4=1$; try to check that $X_1=i$ is a root of said equation.
A: When you speek about solutions of an equation you must always be careful about the "algebraic structure" (In a specialistic language Fields/Rings) where you are searching a solution.
For example, searching a solution in $\mathbb{Z}$ is not equal to searching a solution over $\mathbb{Q}$ even in very simple cases.
An interesting example of this phenomenon is the polynomial $2x-1$.
The related equation has no solution in the ring $\mathbb{Z}$ but a unique solution in $\mathbb{Q}$ (that is in specialistic language $\mathbb{Z}$'s fraction field).
Looking more closely to your case, given a polynomial with just one variable (a so-called "univariate polynomial") of degree $n$, its associated equation has  always $n$ solution in (the field) $\mathbb{C}$ but it can have no solution in -(the field) $\mathbb{R}$
Look for example to the polynomial $x^2+1$.
Algebra in some sense is the study of this differences, but something sure are the following theorems:

Theorem Let $p(x)$ be a univariate polynomial of degree $n$. Then there always exist a "bigger" algebraic structure (called the splitting field of the polynomial p) where the equation $p(x)=0$ has exactely $n$ solutions (counted with multiplicity)

For example, in a case above, the field $\mathbb{C}$ is the splitting field of $x^2+1$ over $\mathbb{R}$
Well, to conclude, let's speack about $\mathbb{C}$.
It is a very "beautifull" algebraic structure because has a very nice property:

Theorem Let $p(x)$ an univariate polynomial of degree $n$ with coefficients in $\mathbb{C}$. Then $p(x)=0$ has exactely $n$ solution over $\mathbb{C}$ (counted with multiplicity)

In particular, since $\mathbb{R} \subset \mathbb{C}$, a real polynomial is a complex polinomial and its associated equation has always $n$ complex solution but cannot have $n$ real solutions.
This is precisely the case of $x^2+1$.
There are "not many" algebraic structure with the same property of $\mathbb{C}$.
This kind of structures are called "algebrically closed fields" and are one of the most important objects of Algebra.
A: The statement that there are always $n$ solutions is only true if you are working in the field of complex numbers. In the field of the real numbers this is not true, as you have shown.
Please note, that even in the field of complex numbers, a solution can occur multiple times, i.e. $(x-1)^2$ has a double zero at $1$. In that case, you need to define how you count the number of solutions.
A: Notice, we have $$x^n=1$$ $$x=(1)^{1/n}$$ $$=(\cos 0+i\sin 0)^{1/n}$$ $$=(\cos 2k\pi+i\sin2k\pi)^{1/n}$$ $$=\left(\cos \left(\frac{2k\pi}{n}\right)+i\sin\left(\frac{2k\pi}{n}\right)\right)$$ Thus there are $n$ roots which can be determined by setting $k=0, 1, 2, \ldots (n-1)$
