$\lim_{n \rightarrow \infty} \frac{(2n)!}{n^{2n}}$ I am trying to show  $$\lim_{n \rightarrow \infty} \frac{(2n)!}{n^{2n}}$$ 
I tried breaking it down, and got stuck when trying to $\left( \frac{2^{n}n!}{n^{n}} \right)$ goes to 0. 

You definitely can't write $(2 n)!$ as $2^n n!$.
The following bound is easy to prove: $n! \le n^n$ (consider $n! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot n \le n \cdot n \cdot \ldots \cdot n = n^n$), but that isn't enough for you.
A sharper bound can be obtained by considering:
$$
(n!)^2 = (1 \cdot n) \cdot (2 \cdot (n - 1)) \cdot \ldots \cdot (n \cdot 1)
$$
We now want a bound for the factors. Take the function:
$$
f(k) = k (n + 1 - k)
$$
To get the maximum we set $f'(k) = 0$, i.e., $n + 1 - 2k = 0$, which gives $k^* = \frac{n + 1}{2}$ As $f''(k^*) = - 2 < 0$, this is the maximum. So:
$$
\begin{align*}
(n!)^2 &\le \left( \frac{n + 1}{2} \right)^{2 n} \\
n! &\le \left( \frac{n + 1}{2} \right)^n
\end{align*}
$$
This should make the rest of the proof simple.
