I have to conclude if the following problem is convergent or not using the root test or ratio test for positive series. 

$$\frac{(n!)^2}{(2n)!}$$

I've shown it to be convergent using the ratio test, but I've run into a problem when using the root test. I've concluded that $\sqrt[n]{(n!)} \to 1$ but Wolfram Alpha claims that $\sqrt[n]{(n!)} \to \infty$.

Here is my work:

I started by reducing the expression:

$$\frac{(n!)^2}{(2n)!} = \frac{(n!)(n!)}{(2n)!}.$$
We have that
$$(n!) = (n)(n-1)(n-2)\cdot...\cdot 1 $$
$$(2n)! = (2n)(2n-1)\cdot...\cdot 1 $$
$$ = 2(n)(n-1/2)(n-1)(n-3/2)(n-2)\cdot...\cdot 1 $$ we rearrange this 
$$ 2(n)(n-1)(n-2)(n-1/2)(n-3/2)\cdot ...\cdot 1 .$$ It should be clear that this "contains" $(n!).$ So now we have:

$$\frac{(n!)(n!)}{(2n)!} = \frac{(n!)}{(2n-1)(2n-3)\cdot ...\cdot 1 }. $$ I now applied the root test to this:

$$\sqrt[n]{\frac{(n!)}{(2n-1)(2-3)\cdot ...\cdot 1}} = {\frac{\sqrt[n]{(n!)}}{\sqrt[n]{(2n-1)(2-3)\cdot ...\cdot 1}}}. $$

I examined $\sqrt[n]{(n)}$:

$$\sqrt[n]{(n)} = e^{(ln(n)\cdot(1/n))}. $$ $e^x$ is continuous so we examine 

$$\lim\_{n\to \infty}(\frac{ln(n)}{n}) \to \frac{\infty}{\infty}. $$ 
We use L'hopital:
$$\lim\_{n\to \infty}(\frac{ln(n)'}{n'}) = \lim\_{n\to \infty}(\frac{1/n}{1}) = \lim\_{n\to \infty}1/n \to 0.$$ 

So we have 
$$\lim\_{n\to \infty}\sqrt[n]{(n!)} = e^0 = 1.$$ 
We will have the same result given $\sqrt[n]{(n-k)}.$ 

We can now return to $\sqrt[n]{(n!)}$:

$$\sqrt[n]{(n!)} = \sqrt[n]{(n)}\sqrt[n]{(n-1)}\sqrt[n]{(n-2)}\*..\*1$$

$$\lim\_{n\to \infty}\sqrt[n]{(n!)} = 1\cdot1\cdot1\cdot ... \cdot 1 = 1 .$$ I end up concluding the same thing regarding $$\lim\_{n\to \infty}\sqrt[n]{{(2n-1)(2-3)\cdot ... \cdot 1}}.$$

I can't figure out why $\sqrt[n]{(n!)} \to \infty,$ according to Wolfram Alpha. Is it because it becomes an indeterminant form of $1^\infty$ ? Or have I simply made a mistake so that it never becomes $1\cdot 1\cdot 1\cdot ... \cdot 1,$ repeating infinitely?

I've looked at some of the other answers concerning the question of why $\sqrt[n]{(n!)} \to \infty$ it doesn't answer why my specific argument is incorrect, and most of them seem to be using a "trick". Given what is in the book I'm using, these seem needlessly complicated or foreign to the reader. I am simply doing this for fun, the problem is already solved by using the ratio test, so perhaps I am simply underestimating the complexity of this problem.