How is the last "=" true? How can the last equality be true? 

$$
G(u)=\frac{g}{(1+u)^g-1}-\frac1u=\frac{g}{gu + \cdots + u^g} - \frac{1}{u}=\sum_{k=0}^\infty c_ku^k
$$



If you look at the left-hadn side of the equals sign, you have
$$
\begin{align}
\frac{g}{gu + \cdots + u^g} - \frac{1}{u} 
&= \frac{1}{u}\frac{g}{g + {g \choose 2} u + \cdots + u^{g-1}} - \frac{1}{u} \\
&= \frac{1}{u}\frac{1}{1 + \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}} - \frac{1}{u}
\end{align}
$$
If you then expand the first term as a geometric series, you will see that the terms with $\frac{1}{u}$ cancel out, leaving a series of the form $\sum_{k=0}^\infty c_ku^k$ for some undetermined coefficents $c_k$.
Edit: Let's expand on this a little bit. Note that we have (formally!) the expansion
$$
\frac{1}{1 + A} = 1 - A + A^2 - A^3 + \cdots
$$
for any $A$. In particular, if we let $A =  \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}$, then we have
$$
\begin{align}
\frac{1}{u}\frac{1}{1 + \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}} - \frac{1}{u}
&=
\frac{1}{u}\frac{1}{1 + A} - \frac{1}{u} \\
&=
\frac{1}{u}(1 - A + A^2 - A^3 + \cdots) - \frac{1}{u} \\
&=
\frac{1}{u} - \frac{1}{u}A + \frac{1}{u}A^2 - \frac{1}{u}A^3 + \cdots - \frac{1}{u} \\
&=
- \frac{1}{u}A + \frac{1}{u}A^2 - \frac{1}{u}A^3 + \cdots
\end{align}
$$
Since $A = \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}$ this now becomes
$$
- \frac{1}{u}\Big(\frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}\Big) + \frac{1}{u}\Big(\frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}\Big)^2 -  \cdots
$$
which is a series that only contains positive exponent terms in $u$.
