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.

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.
