Q:

If $x^n = f(x)g(x)$ as complex polynomials, must $f,g$ be of the form $x^m$?

If $x^n = f(x)g(x)$ as complex polynomials, must $f,g$ be of the form $x^m$? 
i.e. $x^n = x^mx^l$ where $m+l = n$.
This is quite trivial, but I want to make sure I didn't miss anything.
Attempt: If $f$ wasn't of the form $x^m$, and has degree $d$, the fundamental theorem of algebra says it has $d$ roots counted with multiplicity. One of these roots must be nonzero since $f$ isn't of the form $x^m$. But the left hand side of the equation would be nonzero when evaluated at that root.

A:

You can just use induction. I assume you are looking for monic polynomials $f,g$.
Since $x^n$ vanishes in zero, at least one polynomial between $f$ and $g$ must vanish in zero, hence you have $f= x\cdot f_1$ or $g=x\cdot g_1$ and the problem boils down to proving the statement with $n-1$ in place of $n$. Since the statement holds for $n=1$, we're done.

