$f(z) = \frac{1}{z^2-2z+2}$ - $|f^{n}(0)| \leq n!$ - Cauchy inequality Let the function $f(z) = \frac{1}{z^2-2z+2}$. Show that for each integer $n \geq 0$, we have $|f^{n}(0)| \leq n!$.
Honnestly, I don't know how to do that problem, maybe in using the Cauchy inequality (complex analysis).
Is anyone could help me at this point?
 A: We'll use the following result from geometric series to compute the power series of $f$.

Let $c\neq 0$, the power series of $\frac{1}{z-c}$ around $0$ is $\frac{1}{z-c}=\frac{\frac{-1}{c}}{1-\frac{z}{c}}=\frac{-1}{c} (1+(\frac{z}{c})+(\frac{z}{c})^2+(\frac{z}{c})^3+...)$, radius of convergence is $c$.

$f(z)=\frac{1}{(z-a)(z-b)}$, $a=1+i=\sqrt{2}e^{\pi i/4}$, $b=1-i=\sqrt{2}e^{-\pi i/4}$
$$f(z)\\=\frac{1}{z-a}\frac{1}{z-b}\\=\frac{-1}{a}\frac{-1}{b}(1+(\frac{z}{a})+(\frac{z}{a})^2+(\frac{z}{a})^3+...)(1+(\frac{z}{b})+(\frac{z}{b})^2+(\frac{z}{b})^3+...)\\=\frac{1}{2}(1+(\frac{z}{a})+(\frac{z}{a})^2+(\frac{z}{a})^3+...)(1+(\frac{z}{b})+(\frac{z}{b})^2+(\frac{z}{b})^3+...)$$
Because $\frac{f^{(n)}(0)}{n!}=c_n=$ the coefficient in front of $z^n$ in the expansion of $f(z)$. It's sufficient to prove $|c_n|\leq 1$ for each $n$.
Not hard to see from the above equation $c_n=\frac{1}{2}(b^{-n}+a^{-1}b^{-n+1}+...+a^{-n})=\frac{b^{-n-1}-a^{-n-1}}{2(b^{-1}-a^{-1})}=\frac{1}{2(\sqrt2)^n}\frac{\cos((n+1)\frac{\pi}{4})}{\cos(\frac{\pi}{4})}\leq \frac{1}{2\sqrt{2}^{n-1}}\leq 1$.
