Element of finite order is semisimple and eigenvalues are roots of unity

Statement
Suppose $$k$$ is a field, $$V$$ is a finite-dimensional vector space over $$k$$ and $$g$$ is an element in $$GL(V)$$ such that there exists $$n$$ with $$g^n$$ equal to the identity matrix. Then:


 * All the eigenvalues of $$g$$ over the algebraic closure of $$k$$ are $$n^{th}$$ roots of unity.
 * Suppose that either $$k$$ has characteristic zero or $$n$$ is relatively prime to the characteristic of $$k$$. $$g$$ is semisimple, i.e. it is diagonalizable over the algebraic closure of $$k$$.

Applications

 * Trace of inverse is complex conjugate of trace
 * Characters are cyclotomic integers

Proof
$$g$$ satisfies the polynomial $$x^n - 1$$, hence the minimal polynomial of $$g$$ must divide this polynomial. So, every eigenvalue of $$g$$ must satisfy the polynomial $$x^n - 1$$, hence must be a root of unity.

If $$n$$ is relatively prime to the characteristic, then the polynomial $$x^n - 1$$ has no repeated roots, hence the minimal polynomial of $$g$$ has no repeated roots. So, $$g$$ is semisimple, i.e. it is diagonalizable over the algebraic closure.