Proof of $\mathbb E[\exp( \lambda XY)]=\mathbb E[\exp( \lambda^2 X^2)/2]$, where $X,Y$ are independent standard normal random variables I have a heuristic proof of the fact $\mathbb E[\exp( \lambda XY)]=\mathbb E[\exp( \lambda^2 X^2)/2]$,where $X,Y$ are independent standard normal random variables as follows:
Using the fact that $\mathbb E[\exp( \lambda Z)]=\exp(\lambda^2/2)$, where $Z$ is standard normal.
By conditioning, we have 
$\mathbb E[\exp( \lambda XY)]=\mathbb E (\mathbb E[\exp( \lambda XY)\mid X])$, at this point, one may treat $X$ as a constant (heuristically I take $\mathbb E[\exp( \lambda XY)\mid X]=\exp(\lambda^2 X^2/2) $, but is it true and why is it true? ) and use the formula above to get $E[\exp( \lambda^2 X^2)/2]$. 
But how to make this argument formal and rigorous? Any other proof will also be appreciated!

Write 
$$\begin{align}
E(\exp(\lambda XY)) &= \int \exp(\lambda xy)dP_{(X,Y)}(x,y) \tag 1 \\
&= \int \int \exp(\lambda xy)dP_X(x) dP_Y(y) \tag 2\\
&= \int \left(\int \exp(\lambda xy)dP_Y(y)\right) dP_X(x) \tag 3\\
&= \int E(\exp(\lambda xY)) dP_X(x)\\
&= \int \exp( \lambda^2 x^2)/2\; dP_X(x)\\
&= E(\exp( \lambda^2 X^2)/2)
\end{align}$$
$(1)$: law of the unconscious statistician
$(2)$: independence of $X$ and $Y$
$(3)$: Tonelli's theorem
