Can the integral formula for the inverse Z-transform be derived from Cauchy's integral formula?

Question

The Z-transform of the function $ x[n] $ is given by $$ \sum_{n=0}^{\infty} x[n] z^{-n} = X(z) $$ The inverse transform of $ X(z) $ is given by $$ x[n] = \frac{1}{2\pi i} \oint_{\gamma} X(z) z^{n-1} \mathrm{d}z $$ but rearranging Cauchy's formula from $$ f^{(k)}(a) = \frac{k!}{2 \pi i} \oint_{\gamma} \frac{f(z)}{{(z-a)}^{k+1}} \mathrm{d}z $$ to $$ \frac{f^{(k)}(0)}{k!} = \frac{1}{2 \pi i} \oint_{\gamma} \frac{f(z)}{{(z-0)}^{k+1}} \mathrm{d}z $$ then letting $ f(z) = X(z) $ and $ k=-n $ we get $$ \frac{X^{(-n)}(0)}{(-n)!} = \frac{1}{2\pi i} \oint_{\gamma} X(z) z^{n-1} \mathrm{d}z = x[n] $$ It then seems to follow that $$ \sum_{n=0}^{\infty} \frac{X^{(-n)}(0) z^{-n}}{(-n)!} = X(z) $$ Now this is almost the Taylor series of $ X(z) $ save for the indicies decending from $ 0 $ to $ - \infty $. So it is at this point that I am lost. Have I made a mistake?