Constructing equivalent martingale measure (lognormal distribution) Please, help me with next problem. It's from Follmer's and Schied's Stochastic Finance. An introduction in discrete time.

First, I'm looking for conditions for $X_t$ to be a martingale: 
$$
E[X_t|\mathcal{F}_{t-1}] = E[X_0 \prod_{i=1}^{t} e^{\sigma_i Z_i + m_i}|\mathcal{F}_{t-1}] = X_{t-1}\cdot E[e^{\sigma_t Z_t + m_t}] = X_{t-1}\cdot e^{m_t + \frac{\sigma_t^{2}}{2}}
$$
where the last (unconditional) expectation was taken w.r.t. a standard normal density.  
Under a new measure $ P^{*}$, $E^{*}[e^{\sigma_t Z_t + m_t}] = E[\frac{dP_{t}^*}{dP_t} \cdot e^{\sigma_t Z_t + m_t}]$ must be equal to 1 of $m_t + \frac{\sigma_t^2}{2} = 0 \implies m_t = -\frac{\sigma_t^2}{2}$. I want ot find $\frac{dP^*}{dP}$ using two densities (or measures), but I confused, which to take.
Am I on the right way and how to proceed?
I decided to look for a Radon-Nikodym derivative $\frac{dP_{t}^*}{dP_t}$ of a one step, and then take a product of such densities. 
Thanks in advance!
 A: I know this comes late, but here is a possible solution:
As you already indicated, it might be reasonable guess, that $dP^*/dP$ has some kind of product structure; i.e. there are $\sigma(Z_t)$-m.b. random variables $Y_t$ s.t. $$dP^*_t/dP=\prod_{i=1}^tY_i\,.$$ With this ansatz we get by Prop. A.16:
\begin{align*}
 E^*(X_t\mid X_{t-1})
&=\frac{1}{E(dP^*_t/dP\mid X_{t-1})}E(X_tdP^*_t/dP\mid X_{t-1})\\
&=\frac{1}{\left(\prod_{i=1}^{t-1}Y_i\right)E(Y_t\mid X_{t-1})}\left(\prod_{i=1}^{t-1}Y_i\right)X_{t-1}E(e^{\sigma_tZ_t+m_t}Y_t\mid X_{t-1})\\
&=\frac{1}{E(Y_t)}X_{t-1}E(e^{\sigma_tZ_t+m_t}Y_t)\,.
\end{align*}
Further we know that in order to be a density, we have by induction on $i$ and the independence of the $Y_i$ that $E(Y_i)\overset{!}{=}1$ for each $i$. Thus we have to choose $Y_i$ in such a way that
$$ E(Y_i)=1 \quad \text{and}\quad E(e^{\sigma_iZ_i+m_i}Y_i)=1\,.$$
Moreover we have the restriction that $dP^*/dPX_i$ is supposed to be log-normal distributed.
One can easily check that by the formula for the expected value of a log-normal r.v. the following does the job:
$$ Y_i=\exp\left(-(\sigma_i/2+m_i/\sigma_i)Z_i-(\sigma_i/2+m_i/\sigma_i)^2/2\right)\,.$$
