If we know a random variable $X$ satisfies
$$E(X^k) = \left\{ \begin{array}{lll} 1 \cdot 3\cdot ...\cdot k & \text{if} & k \text{ is even}\\ 0 & \text{if} & k \text{ is odd}\\ \end{array} \right. $$
how to prove it follows $N(0,1)$?
Attempted:
If the moment generating function could be proved to be $e^{t^2/2}$ then we are done. But how to find its MGF from the given property?
Remember that the Taylor Series of $e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}$, so we have the expansion of $e^{t^2/2} = \sum_{n=0}^{\infty}\frac{t^{2n}}{2^nn!}$.
But the MFG of $X$ can write as follows
$$E[e^{tX}]=1+tE[X]+\frac{t^2E[X^2]}{2!}+\frac{t^3E[X^3]}{3!}+\frac{t^4E[X^4]}{4!}+...$$
Observe that all the odd powers of $t$ will vanish because $E[X^k] = 0$ when $k$ is odd. Now to the case when $k$ is even I didn't understand the expression for $E[X^{k}]$. Maybe there is some typo. But I believe it is just a question of plug the right expression and to compare $\frac{E[X^{2n}]}{(2n)!}$ with $\frac{1}{2^nn!}$...
Edit: I guess you meant $E[X^k] = 1\cdot 3 \cdot ... \cdot k-1$, i.e., the product of all odd numbers less than $k$. Observe that, writing $k=2n$,
$$\frac{1\cdot 3 \cdot ... \cdot (2n-1)}{(2n)!}=\frac{1}{2\cdot 4 \cdot ... \cdot 2n}=\frac{1}{2^nn!}$$ that it is exactly what you want. (Just extract all the factor of 2 in each number of the product $2\cdot 4 \cdot ... \cdot 2n$. You have $n$ two's...)