complex induction zero points holomorphic function Let $h:G\rightarrow \mathbb{C}$ be a holomorph function and $z_{0}$ a point in $G$ with $h(z_{0}) \ne 0$ . Let m be a natural number. Show that the function : $f:G\rightarrow \mathbb{C}$ defined by $f(z)=(z-z_{0})^{m}h(z)$ at $z_{0}$ ist a zero point of order m and that it holds that : $f^{(m)}(z_{0}) = m!h(z_{0})$
This is my proof: 
 Induction hypothesis  : $f^{(m)}(z_{0}) = m!h(z_{0})$ 
 Induction beginning  : $m=0 \rightarrow f^{(0)}(z_{0}) = h(z_{0})$
 Induction step  : $m \rightarrow m+1 $: $f^{(m+1)}(z_{0})= ((z-z_{0})^{m+1}h(z_{0}))^{(m+1)} = h(z_{0})^{(m+1)}(z_{0}-z_{0})^{m+1} + (m+1)!h(z_{0}) = (m+1)!h(z_{0})$
Is this proof correct. Tell me please. 
 A: Another way to prove it is to use Cauchy integral formula: since $h(z)$ is holomorphic in $G$,  $f(z)=(z-z_{0})^{m}h(z)$ is also holomorphic in $G$. For $r>0$ sufficiently small such that $\{z:|z-z_0|\leq r\}\subset G$,  by Cauchy integral formula we have
$$f^{(m-1)}(z_0)=\frac{(m-1)!}{2\pi i}\int_{|z-z_0|=r}\frac{f(z)}{(z-z_0)^{m}}dz=\frac{(m-1)!}{2\pi i}\int_{|z-z_0|=r}\frac{h(z)(z-z_{0})^{m}}{(z-z_0)^{m}}d
z$$$$=\frac{m!}{2\pi i}\int_{|z-z_0|=r}h(z)dz=0.$$
Here the last equality follows from the fact that $h(z)$ is holomorphic. By Cauchy integral formula once again, we have
$$f^{(m)}(z_0)=\frac{m!}{2\pi i}\int_{|z-z_0|=r}\frac{f(z)}{(z-z_0)^{m+1}}dz=\frac{m!}{2\pi i}\int_{|z-z_0|=r}\frac{h(z)(z-z_{0})^{m}}{(z-z_0)^{m+1}}d
z$$$$=\frac{m!}{2\pi i}\int_{|z-z_0|=r}\frac{h(z)}{z-z_0}dz=m!h(z_0)\neq0.$$
This shows that $z_0$ is a zero of order $m$ to $f(z)$, and $f^{(m)}(z_0)=m!h(z_0)$, as required.
A: Georges Elencwajg has already pointed out in the comments how you could correct your induction attempt.
It would be best to include your definition of the order of a zero, because this essentially is the definition according to some.  Alternatively, $f$ has a zero of order $m$ at $z_0$ if $f^{(k)}(z_0)=0$ when $k\in\{0,1,\ldots,m-1\}$, and $f^{(m)}(z_0)\neq 0$.  You can find $f^{(k)}(z)$ using the product rule.  First you could note that when $0\leq k\leq m$, the $k^\text{th}$ derivative of $(z-z_0)^m$ is $\frac{m!}{(m-k)!}(z-z_0)^{m-k}$.  Therefore
$$f^{(k)}(z)=\sum_{n=0}^k{k \choose n}\frac{m!}{(m-k+n)!}(z-z_0)^{m-k+n}h^{(n)}(z).$$
When $k<m$, the exponent of $(z-z_0)$ is positive in each summand, so $f^{k}(z_0)=0$.  When $k=m$, the only summand without a positive exponent on $(z-z_0)$ is where $n=0$, so we get $f^{(m)}(z_0)={m \choose 0}\frac{m!}{0!}h(z_0)=m!h(z_0).$  Since $h(z_0)\neq 0$, this also shows that the zero is of order $m$.  
Although no explicit induction appears in this answer, both the product rule for higher derivatives and the formula for the higher derivatives of $(z-z_0)^m$ could be proven by induction.  
$_{\text{(For simplicity I am using the notational convention }(\text{anything})^0=1\text{.)}}$
