expansion of a generating function I found this formula in a book
$$\sqrt{1-4x}=1-2\sum_{n=1}^{\infty}\frac{1}{n} {{2n-2}\choose {n-1}} x^n$$
How can I prove that?

There are quite a few ways to prove it. 
You could note that $\frac1n\binom{2n-2}{n-1}$ is $C_{n-1}$, the $(n-1)$-st Catalan number, so that the formula can be written
$$\sqrt{1-4x}=1-2\sum_{n\ge 0}\frac1{n+1}\binom{2n}nx^{n+1}=1-2\sum_{n\ge 0}C_nx^{n+1}\;.$$
Now you can simply manipulate the known generating function for the Catalan numbers:
$$\sum_{n\ge 0}C_nx^n=\frac{1-\sqrt{1-4x}}{2x}\;,$$
so
$$\begin{align*}
\sqrt{1-4x}&=1-2x\sum_{n\ge 0}C_nx^n\\
&=1-2\sum_{n\ge 0}C_nx^{n+1}\;,
\end{align*}$$
as desired.
Or you can use the generalized binomial theorem:
$$\begin{align*}
(1-4x)^{1/2}&=\sum_{n\ge 0}\binom{1/2}n(-4x)^n\\
&=1+\sum_{n\ge 1}\binom{1/2}n(-1)^n4^nx^n\\
&\overset{*}=1+\sum_{n\ge 1}\frac{(-1)^{n-1}}{2^{2n-1}n}\binom{2n-2}{n-1}(-1)^n4^nx^n\\
&=1-\sum_{n\ge 1}\frac{2^{2n}}{2^{2n-1}n}\binom{2n-2}{n-1}x^n\\
&=1-2\sum_{n\ge 1}\frac1n\binom{2n-2}{n-1}x^n\;,
\end{align*}$$
where the starred step is carried out in detail here.
