Proof/Derivation of Closed form of Binomial Expression $\sum\limits_{k=0}^{2n}(-1)^k\binom{2n}{k}^2$ The binomial expression given as follows:
$$\sum_{k=0}^{2n}\left(-1\right)^{k}\binom{2n}{k}^{2}$$
results nicely into the following closed form:
$$(-1)^{n}\binom{2n}{n}$$
I wish to know how exactly is it being done? I haven't been able to make much progress in solving it.
My approach:
\begin{align}
\sum_{k=0}^{2n}(-1)^{k}\binom{2n}{k}^{2} = \binom{2n}{0}^{2} - \binom{2n}{1}^{2} + \binom{2n}{2}^{2} - ... -\binom{2n}{2n-1}^{2} + \binom{2n}{2n}^{2} \\
= \binom{2n}{0}.\binom{2n}{0} - \binom{2n}{1}.\binom{2n}{1} + \binom{2n}{2}.\binom{2n}{2} - ... -\binom{2n}{2n-1}.\binom{2n}{2n-1} + \binom{2n}{2n}.\binom{2n}{2n} \\
\text{By Symmetry of binomial coefficients} \\
= \binom{2n}{2n}.\binom{2n}{0} - \binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n-2}.\binom{2n}{2} - ... -\binom{2n}{1}.\binom{2n}{2n-1} + \binom{2n}{0}.\binom{2n}{2n} \\
= \binom{2n}{2n}.\binom{2n}{0} + \binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n-2}.\binom{2n}{2} + ... +\binom{2n}{1}.\binom{2n}{2n-1} + \binom{2n}{0}.\binom{2n}{2n} - 2.\left(\binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n -3}.\binom{2n}{3} + ... + \binom{2n}{1}.\binom{2n}{2n-1}\right) \\
\text{By Vandermond's identity, first component, i.e.not enclosed within -2.(...) evaluates to C(4n, 2n)} \\
\binom{4n}{2n} - 2.(\binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n -3}.\binom{2n}{3} + ... + \binom{2n}{1}.\binom{2n}{2n-1})
\end{align}
I'm lost beyond this point. It will be extremely helpful if someone can direct me in the right direction or provide the answer to this perplexing and challenging problem. Thank you.
 A: I will use the notation $[x^k]\,f(x)$ for denoting the coefficient of $x^k$ in the Taylor/Laurent expansion of $f(x)$ around the origin. We have:
$$ S(n)=\sum_{k=0}^{2n}(-1)^k \binom{2n}{k}^2 = \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}\binom{2n}{2n-k}=\sum_{\substack{a,b\geq 0 \\ a+b=2n}}(-1)^a\binom{2n}{a}\binom{2n}{b} $$
and since
$$ \sum_{c\geq 0}(-1)^c \binom{2n}{c} x^c = (1-x)^{2n}, \qquad \sum_{d\geq 0}\binom{2n}{d} x^d = (1+x)^{2n} $$
it follows that:
$$ S(n) = [x^{2n}] (1-x)^{2n}(1+x)^{2n} = [x^{2n}](1-x^2)^{2n} \stackrel{x^2\mapsto z}{=} [z^n](1-z)^{2n}$$
so $S(n) =\color{red}{ (-1)^n \binom{2n}{n}}$ just follows from the binomial theorem.
A: I'll present a proof using the "Description, Involution, Exception" technique of A. Benjamin and J. Quinn to simplify the left side. Consider counting pairs $(A, B)_k$, where each item is a size-$k$ subset of $\{1, 2, \dots, 2n\}$. Clearly for each $k$ the total number of such pairs is $\binom{2n}{k}^2$. 
Our involution, then, is to take the smallest element either in both sets or neither set, and either remove it from both or add it to both respectively. It should be clear that this is an involution, and that an element and its image are always counted at different parities in our summation, and thus contribute 0 to the total sum.
So, the elements not cancelled by this involution are exactly those pairs which are disjoint and whose union is all $2n$ elements. Being disjoint means having size no more than $n$, by pigeonhole, and totaling $2n$ means having at least $n$ each. Thus this is exactly the $n$-element pairs that have no overlap, which can be counted by choosing $n$ elements for the left item (as the right item must have exactly the rest). Meanwhile, the sign is inherited from the larger summation, where it is based on the parity of $k$, here $n$. This gives us $(-1)^n\binom{2n}{n}$.
A: Think generating functions. The sum is the $x^n$ coefficient of
$$\sum_{k=0}^n\binom nkx^k\sum_{l=0}^n(-1)^l\binom nlx^l.$$
