I'm trying to solve the following problem:
Show that the m.g.f. of the gamma distribution $\Gamma(\alpha,\gamma)$ is $(1-\gamma t)^{-\alpha}$, $t<\gamma^{-1}$, using Theorem 2.1(ii).
This theorem states that:
Theorem 2.1.: Let $\cal{P}$ be a ${natural}$ exponential family given by $f_\eta(\omega)=exp\{\eta^T T(\omega)-\zeta(\eta) \}h(\omega)$, where $\eta$ are the canonical parameters:
(ii) If $\eta_0$ is an interior point of the natural parameter space, then the m.g.f. $\psi_{\eta_0}$ of $P_{\eta_0}oT^{-1}$ is finite in a neighborhood of 0 and is given by:
$\psi_{\eta_0}(t) = exp\{\zeta(\eta_0 +t) - \zeta(\eta_0) \} $
My attempt:
First, I've tried to put the pdf in exponential family and find canonical parameters:
$f_x = \frac{1}{\Gamma(\alpha)\gamma^\alpha}x^{\alpha-1}e^{-\frac{x}{\gamma}}=exp\{\alpha \, logx - x \frac{1}{\gamma} - log\Gamma(\alpha)-\alpha\,log\gamma \}$
So that:
$\eta = [\alpha,\frac{1}{\gamma}] = [\eta_1,\eta_2]$ and $T = [logx,-x]$ (I know it can be selected differently)
$\zeta_{\eta} = log\Gamma(\eta_1)-\eta_1\,log\eta_2$
Now I got stuck, since:
(i) How can I get $\zeta(\eta_0 +t) $ since $\eta$ is a vector and I cant sum t to it?
(ii) How to get rid of the $\Gamma(.)$ in this equation?
The log-partition function, your $\zeta(\cdot)$, can be used to get the moment generating function of the sufficient statistic which is vector valued. So to answer your question $(i)$ you don't add a scalar $t$ but a vector so the moment generating function for the sufficient statistic $T = T(x)$ is given by $$ \mathbb{E}\left[e^{\mathbf{s}\cdot T(x)}\right], $$ where '$\cdot$' is the inner product, and $\mathbf{s}$ is a vector equal in dimension to that of the sufficient statistic. In your particular case we have sufficient statistic $$ \begin{align} T = \begin{bmatrix} \log x \\ -x\end{bmatrix}. \end{align} $$ So $$ \begin{align} \mathbb{E}\left[e^{t X} \right] &= \mathbb{E}\left[ \exp\left\{ \begin{bmatrix} 0 & -t \end{bmatrix}\cdot\begin{bmatrix}\log X \\ -X \end{bmatrix} \right\} \right], \end{align} $$ where our chosen vector $\mathbf{s}$ is taken to be $\mathbf{s} = [0, -t]^T$ and we can use the log-partition function you have calculated where $\zeta(\boldsymbol{\eta}+\mathbf{s}) - \zeta(\mathbf{s})$ is given by $$ \begin{align} \zeta(\eta_1,\eta_2-t)-\zeta(\eta_1,\eta_2) &= \eta_1\log \frac{\eta_2}{\eta_2-t} \\ &= \eta_1\log \frac{1}{1-t\eta_2^{-1}} \\ &= \alpha \log \frac{1}{1-t\gamma}, \end{align} $$ and so finally taking the exponential of this $$ \begin{align} \mathbb{E}\left[e^{tX}\right] = (1-t\gamma)^{-\alpha}. \end{align} $$
To be more explicit about your point $(ii)$ we don't have to worry about the $\Gamma(\cdot)$ function because we are only interested in the mgf of the second component of the sufficient statistic and so only vary the corresponding natural parameters in the log-partition function.
As a final comment I have tried to stick to your notation and chosen parameterisations but watch out for any little mistakes.