I wish to show that if for integers $x,y$
$$ x = y^n$$
if the total number of factors of $x$ is $1 \mod n$. When $n=2$ this reduces to showing that square numbers have an odd number of factors, this is argued by claiming any factor $f$ of a number $X$ has a complement $g$ such that $fg = X$. Therefore if one looks at all the factor pairs of $X$, one will find 2 pairs for every $f,g$ that are unequal. Squares carry a factor $n$ such that $n^2=X$. Thus square numbers will only receive a single pair from this factor, resulting in an odd number of factors.
Now how to easily generalize this to $n=3, ... $