Prove that if $n\geq1$ then $\binom{2n}{2}=$ Prove that if $n\geq1$ then $$\binom{2n}{n}=\sum_{k=0}^{n}(\binom{n}{k})^2$$

This is what I have so far: 
By the Binomial Theorem: $$\binom{2n}{n}=\frac{(2n)!}{(2n-2)!n!}=\frac{2n!~}{n^2(n-1)^2(n-2)^2...1}$$
By the definition of summation: $$\sum_{k=0}^{n}(\binom{n}{k})^2=(\binom{n}{0})^2+(\binom{n}{1})^2+(\binom{n}{2})^2+(\binom{n}{3})^2+...+(\binom{n}{n})^2$$
By the Binomial Theorem => $$\sum_{k=0}^{n}(\binom{n}{k})^2=1^2+(\frac{n!}{(n-1)!})^2+(\frac{n!}{(n-2)!2!})^2+(\frac{n!}{(n-3)!3!})^2+...+1^2$$
$$=1+\frac{n!n!}{((n-1)!)^2}+\frac{n!n!}{((n-2)!)^2(2!)^2}+\frac{n!n!}{((n-3)!)^2(3!)^2}+...+1$$
This is where I get stuck. I'm not sure how to continue from here. I know that what I am getting is starting to look a lot like what I want it to look like but I don't know how to finish, nor am I sure that I've done it correctly to this point. Help? 
 A: Although this question has been answered, I have another proof.
Let $N=\{1,\ldots,n\}$, $N_2=\{1,2\}\times N$, $A$ be the set of subsets of $N$, $B=\{(X,Y)\in A\times A:|X|=|Y|\}$ and $C=\{X\subset N_2:|X|=n\}$.
It's easy to show that
$$|B|=\sum_{k=0}^n\binom nk^2$$
and
$$|C|=\binom {2n}n$$
Therefore, our goal is to show that $|B|=|C|$, so let's build a bijective function from $C$ to $B$.
Let be $X\subset N_2$. $X$ has $k$ elements whose first component is $1$ and $n-k$ elements whose first component is $2$. Define
$$f(X)=\Big(\{v:(1,v)\in X\},\{v:(2,v)\notin X\}\Big)$$
Since $f$ is bijective, we are finished.
To say it in plain words, we can render $N_2$ as a matrix like this:
$$\begin{pmatrix}1&2&\cdots& n\\1&2&\cdots& n\end{pmatrix}$$
Choosing a set $X\subset C$ is choosing $n$ terms of the matrix. This leaves $k$ terms selected in the first row and $k$ terms unselected in the second row. This is like selecting two subsets of the sime size from $N$, which is like selecting an element from $B$.
