We have a fixed $a \in \mathbb{R}_{0}^{+}$, and we know, that $$\Phi_{j} \sim \mathcal{N}(\phi_{j}, s_{j}^{2})$$ for $s_{j}>0$. $\Phi_{j}$ are independent for $j=0,...,J-1$.
I have to proof, that $$ \mathbb{E} \left[ \prod_{j=0}^{J-1} \exp{ \left\{ a \;\Phi_{j} - a\;\phi_{j} - a^{2} \frac{s_{j}^{2}}{2} \right\} } \right] = 1 $$
My first idea was, since $\mathbb{E}[\Phi_{j}]=\phi_{j}$ (1) and $\phi_{j}, s_{j}$ are known, to pull out the deterministic parts and using (1) so that i get $$ \prod_{j=0}^{J-1} \exp \left\{ - \frac{a^{2}s_{j}^{2}}{2} \right\}=1,$$ but that's obviously not purposeful and i am even not sure if that's correct.
I got the hint that this is a moment-generating function. So i can pull out the product (independence) an get $$ \prod_{j=0}^{J-1} \mathbb{E} \left[ \exp{ \left\{ a \;\Phi_{j} - a\;\phi_{j} - a^{2} \frac{s_{j}^{2}}{2} \right\} } \right] = 1 $$ How ist this helpful to get to the result?
$$\mathbb{E} \exp{ \left\{ a \Phi_{j} - a\phi_{j} - a^{2} \frac{s_{j}^{2}}{2} \right\}}=\mathbb E[\exp(a\Phi_j)]\exp(-a\phi_j-a^2s^2_j/2)$$
Note $E[\exp(a\Phi_j)]$ is the m.g.f. for $\Phi_j$. Use the known closed form of this m.g.f.