Show that X has a Chi Square Distribution with 4 degrees of freedom Let $X$ be a random variable such that $E(X^n) = (n+1)!\cdot2^n$ for $n = 1,2,3,\ldots$. 
Show that $X$ has a Chi Square distribution with 4 degrees of freedom. 
I am not sure how to begin this. Can I find the pdf from moments? 

A Chi Square distribution with 4 degrees of freedom has a pdf given by 
$$ f_X(x) = \frac{x e^{-x/2}}{4}$$ and mgf $$M_X(t)= (1-2t)^{-2}.$$
Calculating the moments we have, 
$M^{(1)}_X(0) = (-2) \cdot (1)^{-3} \cdot (-2) = 2 \cdot 2$ 
$M^{(2)}_X(0) = (-2) (-3)  \cdot (1)^{-4} \cdot (-2)(-2) = 3 \cdot 2 \cdot 2^2$ 
$M^{(3)}_X(0) = (-2) (-3) (-4)  \cdot (1)^{-5} \cdot (-2)(-2)(-2) = 4 \cdot 3 \cdot 2 \cdot 2^3$
$M^{(4)}_X(0) = (-2) (-3) (-4) (-5)  \cdot (1)^{-6} \cdot (-2)(-2)(-2)(-2) =5\cdot 4 \cdot 3 \cdot 2 \cdot 2^4.$
Inductively, one can see that the pattern is set. So we can write, 
$M^{(n)}_X(0) = (n+1)!\cdot 2^n$
Since the mgf uniquely determines a distribution, it follows that if the moments are given by $E(X^n) = (n+1)! 2^n$, the distribution is Chi Square with 4 degrees of freedom. 
