How does one show, that the moment generating function of $T=(T_1, \ldots, T_k)$ of an natural parameterized $k-$ dimensional exponential family has the following form:
$$ \psi(t) = E[exp(t^{\top}T)] = \dfrac{C(\theta_1, \ldots, \theta_k)}{C(\theta_1+t_1, \ldots, \theta_k+t_k)}$$
for $t=(t_1, \ldots, t_k) \in \mathbb{R}^k$.
Note that in the natural form, the exponential family is $f(x\mid \theta) = h(x)g(\theta) \exp (\theta' T(x))$ for some appropriate $h,g,T$. Now,
$$\mathbb{E}\left[\exp( t' T)\right] = \int h(x)g(\theta) e^{\theta' T(x)} e^{t'T(x)} \,\mathrm{d}x = g(\theta) \int h(x) e^{(\theta + t)' T(x)} \,\mathrm{d}x$$
But we know that $ f(x| \theta + t) = h(x) g(\theta + t) e^{(\theta + t)'T}$ is a density on $x$, which implies that $\int h(x) g(\theta + t) e^{(\theta + t)'T} \,\mathrm{d}x = 1$, and thus the final integral above must be $1/g(\theta + t)$. You now have the desired form.