This is what I want to prove: If $f$ is a nonzero holomorphic function on the open disc $\mathbb D$ then $f(z)=z^n g(z)$ for some $n \ge 0$ and some $g$ holomorphic on the open disc $\mathbb D$ satisfying $g(0)\ne 0$.
Here's my attempt: Suppose not. Then for every $g\in H(\mathbb D)$ and every $n \ge 0$, if $f(z)=z^n g(z)$ on $\mathbb D$ then $g(0)=0$. Then by appealing to Taylor's theorem, we can prove that $f\equiv 0$ on $\mathbb D$.
I believe my attempt is correct but I was looking for alternative proofs which avoid the use of Taylor's theorem. Is it possible to prove this using the identity theorem?
The one-step Taylor's theorem, also known as the fundamental theorem of calculus, gives you an explicit expression for $g$. A simple recurrence then allows to conclude. $$ f(z) = f(0) + z \int_0^1 f'(tz) \, dt. $$ If $f(0) = 0$, we put $f_1(z) = \int_0^1 f'(tz) \, dt$. It is holomorphic by a standard result on integrals of function depending holomorphically on a parameter. If $f_1(0) \neq 0$, we take $g = f_1$, if not we repeat with $f_1$ in place of $f$. If the recurrence does not end, this means that all derivatives of $f$ at $0$ are equal to $0$, so $f$ is identically $0$.
Interestingly, the argument gives a related result for $C^\infty$ functions on ${\bf R}$. If the nth-1 first derivatives of $f$ at 0 are 0, then there is a $C^\infty$ function $g$ such that $f(x) = x^n g(x)$. The value of $g$ at zero is equal to $f^{(n)}(0)$, which may or may not be zero.