Proof that a Factorial cannot be a Double Factorial I am looking for a proof of the following:

Given an integer $n > 2$, $n$ cannot be both a factorial as well as a double factorial.

In more mathematical terms:
$$ \forall n \in \mathbb{Z} : n > 2,\ \nexists\ (i,j) : n = i!\ and\ n = j!!$$
I have been able to "show" that this is true by running a Python program that simply generates searches for equal factorials and double factorials in increasing order, but have not been able to come up with a sufficient proof myself.
I believe that the following formula for double factorials will help:
$$ 0!! = 1!! = 1,\ n!! = n!/\left((n-1)!!\right)  \forall\ n > 1$$
I think it can be done both by induction and by contradiction. I would love to see both proofs!
 A: If $k! = (2n)!! = 2^nn!$, then $k>n$ and so $$\binom{k}{n} = \dfrac{k!}{n!(k-n)!} = \dfrac{2^n}{(k-n)!}.$$
This forces that $k-n=1$ or $k-n=2$, otherwise the right hand side would not be an integer. If $k=n+1,$ then it means $n+1 = 2^n$, whose only integer solution is $n=1.$ If $k=n+2,$ then $(n+2)(n+1) = 2^n$, which means both $n+2, n+1$ are powers of two. So $n=0$, but it does not satisfy the equation. 
A: If $j=2k$ then $j!!=2^kk!$
and if $2^kk!=n!$ then we have very many problems.
If $j=2k-1$ we also have problems because $j!!$ is odd.
A: Suppose for contradiction we have some $j,n$ such that $j!! = n!$. We split this into $2$ cases.
Suppose $j = 2k$ with $k > 1$, then $j!! = 2^k k! = n!$, so in particular $n > k$ and $2^k = n(n-1)(n-2) \cdots (k+1)$. If $n > k+1$, note at least one of these numbers on the LHS is odd and greater than $1$, so we have an even number ($2^k$) divisible by an odd number ($n, k+1,$ or something in between). This is a contradiction. Likewise if $n = k+1$, we have $2^{k} = k+1$, which implies $k=0$, contradicting $k > 1$. Therefore $j$ cannot be even.
Now, suppose $j = 2k-1$, then $j!!$ is odd, yet $n!$ for any $n > 1$ is even. Hence $j!! \neq n!$.
