I have to show that $E[e^{(a^2/2)N^2}]=E[e^{(aNN')}]$ and tell for which values of $a$ these quantities are finite.
$N$ and $N'$ are independent $\mathcal{N}(0,1)$ random variables
I computed the one on the right side, according to my computations is $\frac{1}{\sqrt{(1-a)}}$, how should I arrange the problem on the left side?
ps just to be sure since it appears while computing the integral on the right side: I arranged the "inner" integral in order to have the density of a $\mathcal{N}(ax,1)$ random variable (I was integrating in the other variable), can I safely assume its integral is equal to $1, \forall x \in R$? I see no reason why this shouldn't hold, but these are the first approaches with multivariate random variables so I'm not really sure.
One knows that, for every $b$, $E(\mathrm e^{bN'})=G(b)$, with $G(b)=\mathrm e^{b^2/2}$ hence, by independence, $E(\mathrm e^{aNN'}\mid N)=E(\mathrm e^{(aN)N'}\mid N)=G(aN)$, which solves the first part.
For the second part, only one random variable is involved, with known distribution, hence a direct computation shows that $E(\mathrm e^{a^2N^2/2})=1/\sqrt{1-a^2}$, for every $|a|\lt1$ (hence the tentative solution in the question has a misprint).