Unable to solve expression for $x$ I'm trying to solve this expression for $x$:
$$\frac{x^n(n(1-x)+1)}{(1-x)^2}=0$$
I'm not sure where to begin (especially getting rid of the $x^n$ part), any hints or tips are appreciated.

$\frac{x^n*(n(1-x)+1)}{(1-x)^2} = 0$ is equivalent to $x^n*(n(1-x)+1) = 0$ and $x \neq 1$.
But $x^n*(n(1-x)+1) = 0$ is equivalent to $x = 0$ or $n(1-x) + 1 = 0$, e.g. $x=0$ or $1+\frac{1}{n}$.
Finally, the solutions are $$x \in \{0, 1 + \frac{1}{n}\}$$
