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
$$E[V^s] = \frac {\alpha +s-1}{\alpha+s}E[V^{s-1}]$$
There's no playing about with series needed, no worrying about integrals (well not directly). This is "think about what MGFs are" kind of question.
Consider the definition of an MGF -- $\quad M_X(t)=E(e^{tX})$.
Given that definition, what should $M_X(0)$ be?
Do you notice the problem?
Best Answer
If the moment generating function $M_X(t) = \mathbb{E} e^{t X}$ of the random variable $X$ exists (for $t$ in some open interval containing zero), then all the moments of $X$ exists. So one way to show that $t$ distributions do not have moment generating functions is to show that not all moments exist.
But it is well known that the $t$-distribution with $\nu$ degrees of freedom only have moments up to order $\nu-1$, so the mgf do not exist.