My working so far: $X_1=X_2=X_3=...=X_n$ are standard normal.
$Y=e^{X_1}e^{X_2}...e^{X_n}=e^{nX}$
I have tried to find the distribution of $Y$ using this method:$$F_Y(y)=Pr(e^{nX}\le y)$$ $$=Pr(X\le \frac{ln(y)}{n})$$ $$=\int_{-\infty}^{\frac{ln(y)}{n}}f_X(x)dx$$ $$\rightarrow f_y(y)=\frac{1}{n y}f(\frac{ln(y)}{n})$$ $$=\frac{1}{ny\sqrt{2\pi}}e^-\frac{ln^2(y)}{2n}$$
Is this correct? I feel that I'm missing something, namely I don't like that $n$ being in there but I don't know how to work this without that $n$ being there.
Additionally, how would I find the expected value and variance of $Y$?
The moment generating function of a Normal distribution with mean $\mu$ and variance $\sigma^2$ is
$$ M_X(t) = e^{t\mu + \sigma^2 t^2/2} $$
So comparing with your problem, you see that all variables $X_i$ are standard independent normal variables, moreover
$$ Y = \prod_{i=1}^ne^{X_i}= e^{\sum_{i=1}^n X_i} = e^{Z} $$
Where $Z\sim \mathcal{N}(0,n)$. So $Y$ is distributed lognormal with parameters $\mu = 0$ and $\sigma = \sqrt{n}$