Since $0 < \delta <1$, then $\log \delta <0$. So the density of $V$ is
$$g(v) = f\left(\frac{\log v}{\log \delta}\right)\cdot \left|\frac{\partial \tau }{\partial V }\right|
= \lambda \exp\left\{-\lambda \frac{\log v}{\log \delta}\right\}\frac{1}{v |\log \delta|} \\= \frac {\lambda}{|\log \delta|}\frac 1v\exp\left\{ \frac{\lambda}{|\log \delta|}\log v\right\}$$
Set $\alpha \equiv \frac{\lambda}{|\log \delta|}$. Then, manipulating,
$$g(v) = \alpha v^{\alpha-1}, \;\;v\in [0,1] $$
which is the density of a $\text{Beta}(\alpha,1)$ distribution.
The moment generating function is
$$MGF_V(\alpha,1,t) = 1 +\sum_{k=1}^{\infty} \left( \prod_{r=0}^{k-1} \frac{\alpha+r}{\alpha+r+1} \right) \frac{t^k}{k!}$$
and which, among other things provides a nice recursive formula
$$E[V^s] = \frac {\alpha +s-1}{\alpha+s}E[V^{s-1}]$$
Let $Z$ have a standard normal distribution, with mean $0$ and variance $1$, then $(Z+\mu)^2$ has a noncentral chi-squared distribution with one degree of freedom. The moment-generating function of $(Z+\mu)^2$ then is
\begin{equation}
E[e^{t(Z+\mu)^2}] = \int_{-\infty}^{+\infty} e^{t(z+\mu)^2} f_Z(z) \text{d}z
\end{equation}
with $f_Z(z) = \frac{1}{\sqrt{2\pi}}e^{-z^2/2}$ the density of $Z$. Then,
\begin{align*}
E[e^{t(Z+\mu)^2}] & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{t(z+\mu)^2} e^{-z^2/2}\text{d}z\\
& = \frac{1}{\sqrt{2\pi}} \int \exp[-(\frac{1}{2}-t)z^2+2\mu t z +\mu^2 t] \text{d} z\\
& = \frac{1}{\sqrt{2\pi}} \int \exp[-(\frac{1}{2}-t) (z-Q)^2+\mu^2 t + \frac{2\mu^2 t^2}{1-2t}] \text{d} z \qquad\text{with } Q=\frac{2\mu t}{1-2t}\\
& = \exp[\mu^2 t + \frac{2\mu^2t^2}{1-2t}] \frac{1}{\sqrt{2\pi}} \int \exp[-\frac{(z-Q)^2}{2(1-2t)^{-1}}] \text{d}z\\
& = \exp[\mu^2 t + \frac{2\mu^2t^2}{1-2t}] (1-2t)^{-1/2} \times \frac{1}{\sqrt{2\pi}(1-2t)^{-1/2}} \int \exp[-\frac{(z-Q)^2}{2(1-2t)^{-1}}] \text{d}z\\
& = \exp[\mu^2 t + \frac{2\mu^2t^2}{1-2t}] (1-2t)^{-1/2} \times 1\\
& = \exp[\frac{\mu^2 t}{1-2t} ] (1-2t)^{-1/2}
\end{align*}
By definition, a noncentral chi-squared distributed random variable $\chi^2_{n,\lambda}$ with $n$ df and parameter $\lambda=\sum_{i=1}^n \mu_i^2$ is the sum of the squares of $n$ independent normal variables $X_i=Z_i+\mu_i$, $i=1,\ldots,n$. That is,
\begin{equation*}
\chi^2_{n,\lambda} = \sum_{i=1}^n X_i^2 = \sum_{i=1}^n (Z_i+\mu_i)^2
\end{equation*}
Because all terms are jointly independent and using the above result, we have
\begin{align*}
E[e^{t\chi^2_{n,\lambda}}] & = E[\prod_i \exp[t(Z_i+\mu_i)^2]]\\
& = \prod_i E[\exp[t(Z_i+\mu_i)^2]]\\
& = \prod_i (1-2t)^{-1/2} \exp[\frac{\mu_i^2t}{1-2t}]\\
& = (1-2t)^{-n/2} \exp[\frac{t}{1-2t}\sum_i \mu_i^2]\\
& = (1-2t)^{-n/2} \exp[\frac{\lambda t}{1-2t}]\\
\end{align*}
Best Answer
Yes, since $\chi^2$ is a sum of $Z_i^2$ the MGF is a product of individual summands. But then you need the MGF of $Z_i^2$ which is $\chi^2$ with 1 degree of freedom. The obvious way of calculating the MGF of $\chi^2$ is by integrating. It is not that hard:
$$Ee^{tX}=\frac{1}{2^{k/2}\Gamma(k/2)}\int_0^\infty x^{k/2-1}e^{-x(1/2-t)}dx$$
Now do the change of variables $y=x(1/2-t)$, then note that you get Gamma function and the result is yours. If you want deeper insights (if there are any) try asking at http://math.stackexchange.com.