<s>[INST] Binomial identity $\binom{0}{0}\binom{2n}{n}+\binom{2}{1}\binom{2n-2}{n-1}+\binom{4}{2}\binom{2n-4}{n-2}+\cdots+\binom{2n}{n}\binom{0}{0}=4^n.$ Prove the  identity
$$\binom{0}{0}\binom{2n}{n}+\binom{2}{1}\binom{2n-2}{n-1}+\binom{4}{2}\binom{2n-4}{n-2}+\cdots+\binom{2n}{n}\binom{0}{0}=4^n.$$
This is reminiscent of the identity $\sum_{i=0}^n\binom{n}{i}^2=\binom{2n}{n}$, which has a nice combinatorial interpretation of choosing $n$ from $2n$ objects. But the identity in question is not easily related to a combinatorial interpretation. Also, for an induction proof, it is not clear how to relate the identity with $n$ to $n+1$.
 [/INST] By  way of  enrichment here  is  another algebraic  proof using  basic
complex variables.  As I pointed out  in the comment  this identity is
very simple using a convolution,  so what follows should be considered
a learning exercise.

We seek to compute
$$\sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k}.$$
Introduce the integral representation
$${2n-2k\choose n-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-2k}}{z^{n-k+1}} \; dz.$$
We use  this to obtain an integral  for the sum. Note  that when $k>n$
the  pole  at  zero  disappears  which  means  that  the  integral  is
zero. Therefore we may extend the sum to infinity, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{2n}}{z^{n+1}} 
\sum_{k\ge 0} {2k\choose k} \frac{z^k}{(1+z)^{2k}}\; dz.$$
Recall that
$$\sum_{k\ge 0} {2k\choose k} w^k = \frac{1}{\sqrt{1-4w}}$$
so this becomes
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{2n}}{z^{n+1}} \frac{1}{\sqrt{1-4z/(1+z)^2}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{2n}}{z^{n+1}} \frac{1+z}{\sqrt{(1+z)^2-4z}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{2n+1}}{z^{n+1}} \frac{1}{1-z} \; dz.$$
Extracting coefficients we get
$$\sum_{q=0}^n {2n+1\choose q} = 
\frac{1}{2} 2^{2n+1} = 2^{2n} = 4^n.$$ 

Apparently this method is due to Egorychev.

Addendum.
The Lagrange inversion proof goes like this. We seek to compute
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{k+1}} \frac{1}{\sqrt{1-4z}} \; dz.$$
Put $1-4z = w^2$ so that 
$z= \frac{1}{4} - \frac{1}{4} w^2$ and $dz = -\frac{1}{2} w \;  dw$
to get
$$-\frac{1}{2\pi i}
\int_{|w-1|=\epsilon} 
\frac{4^{k+1}}{(1-w^2)^{k+1}} \frac{1}{w} \frac{1}{2} w \; dw
\\ = -\frac{2^{2k+1}}{2\pi i}
\int_{|w-1|=\epsilon} 
\frac{1}{(1-w)^{k+1}} \frac{1}{(1+w)^{k+1}} \; dw
\\ = -\frac{2^{2k+1}}{2\pi i}
\int_{|w-1|=\epsilon} 
\frac{1}{(1-w)^{k+1}} \frac{1}{(2+w-1)^{k+1}} \; dw
\\ = -\frac{2^k}{2\pi i}
\int_{|w-1|=\epsilon} 
\frac{1}{(1-w)^{k+1}} \frac{1}{(1+(w-1)/2)^{k+1}} \; dw
\\ = -\frac{2^k (-1)^{k+1}}{2\pi i}
\int_{|w-1|=\epsilon} 
\frac{1}{(w-1)^{k+1}} \frac{1}{(1+(w-1)/2)^{k+1}} \; dw.$$
Extracting coefficients we obtain
$$- 2^k (-1)^{k+1} {k+k\choose k} \frac{(-1)^k}{2^k}
= {2k\choose k}.$$
It is not difficult to see that in the above substitution the image of
a small  radius counterclockwise circle  around the origin in  the $z$ 
plane  is  a small   radius   circle   around   $w=1$  also  traversed
counterclockwise.
A similar calculation may be found at this 
MSE link.
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