Prove that $\oint_{\gamma}^{}{\frac{f'(z)}{f(z)}dz}= -2\pi iInd_{\gamma}(x)$ I have the following statement but I don't see exactly why it's true. So if anyone could help it would be a lot appreciated :

Let $x \in U  $ and $f$ a holomorphic function on $U$\$x$

$f(z) \neq 0 $ ,  $\forall z\in U$\ $x$

$f$ has a pole of order $n\geq 1$

So we have : 

$$\oint_{\gamma}^{}{\frac{f'(z)}{f(z)}dz}= -2\pi  niInd_{\gamma}(x)$$

EDIT : if $\gamma$ is a loop in $U$\ $x$ we have the statement

I don't see how he jumped to that conclusion ..
That's a particular case of the argument principle. In general, $\frac1{2\pi i}\oint_\gamma\frac{f'(z)}{f(z)}\,\mathrm dz$ is the number of zeros of $f$ minus the number of poles (counted with their multiplicities). Since, in this case, there are $0$ zeros and one pole with multiplicith $n$, $\frac1{2\pi i}\oint_\gamma\frac{f'(z)}{f(z)}\,\mathrm dz=-n$.
That's a particular case of the argument principle. In general, $\frac1{2\pi i}\oint_\gamma\frac{f'(z)}{f(z)}\,\mathrm dz$ is the number of zeros of $f$ minus the number of poles (counted with their multiplicities). Since, in this case, there are $0$ zeros and one pole with multiplicith $n$, $\frac1{2\pi i}\oint_\gamma\frac{f'(z)}{f(z)}\,\mathrm dz=-n$.