The exponential generating function for the central binomial coefficients

Extracting the coefficient we seek to show

$${2n\choose n} \frac{1}{n!} = [z^n] \sum_{k\ge 0} \frac{2^k}{k!} z^k \sum_{k\ge 0} \frac{1}{k! \times k!} z^{2k}.$$

This is

$${2n\choose n} \frac{1}{n!} = \sum_{q=0}^{\lfloor n/2 \rfloor} \frac{1}{q! \times q!} \frac{2^{n-2q}}{(n-2q)!}.$$

or

$${2n\choose n} = \sum_{q=0}^{\lfloor n/2 \rfloor} \frac{1}{q! \times q!} \frac{2^{n-2q} \times n!}{(n-2q)!}.$$

or

$${2n\choose n} = \sum_{q=0}^{\lfloor n/2 \rfloor} \frac{(n-q)!}{q! \times q!} \frac{2^{n-2q} \times n!}{(n-q)! \times (n-2q)!}.$$

In terms of binomial coefficients

$${2n\choose n} = \sum_{q=0}^{\lfloor n/2 \rfloor} {n\choose q} {n-q\choose q} 2^{n-2q}.$$

The RHS is

$$\sum_{q=0}^{\lfloor n/2 \rfloor} {n\choose q} {n-q\choose n-2q} 2^{n-2q} \\ = 2^n [z^n] (1+z)^n \sum_{q=0}^{\lfloor n/2 \rfloor} {n\choose q} 2^{-2q} (1+z)^{-q} z^{2q}.$$

The coefficient extractor combined with the $z^{2q}$ factor enforces the upper limit and we may write

$$2^n [z^n] (1+z)^n \sum_{q\ge 0} {n\choose q} 2^{-2q} (1+z)^{-q} z^{2q} \\ = 2^n [z^n] (1+z)^n \left(1+\frac{z^2}{4(1+z)}\right)^n \\ = 2^n [z^n] (1+z)^n \frac{(4+4z+z^2)^n}{4^n (1+z)^n} \\ = 2^{-n} [z^n] (z+2)^{2n} \\ = 2^{-n} {2n\choose n} 2^n = {2n\choose n}.$$

This is the claim.

There are two facts I am going to use in this proof:

$\binom{2n}{n}=\frac{2^{2n}}{\pi}\int\limits_0^1\frac{y^n}{\sqrt{y(1-y)}}dy\tag1$

And the modified Bessel function of the first kind with zero order can be expressed as:

$I_0(x)=\frac{1}{\pi}\int\limits_0^\pi e^{x cos(\theta)} d\theta = \frac{1}{\pi} \int\limits_{-1}^{1} \frac{e^{xt}}{\sqrt{1-t^2}}dt\tag2$ where $t=cos\theta$

First put (1) into the LHS of the statement and replace the order of the summation and integration:

$\frac{1}{\pi }\int\limits_0^1 \frac{1}{\sqrt{y(1-y)}}\sum\limits_{n=0}^\infty\frac{(4yx)^n}{n!}dy\tag3$

Performing the summation in (3) we get:

$\frac{1}{\pi}\int\limits_0^1 \dfrac{e^{4xy}}{\sqrt{y(1-y)}}dy\tag4$

Using the following substution: $y=r+\frac{1}{2}$ we have:

$\frac{2e^{2x}}{\pi}\int\limits_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{e^{4xr}}{\sqrt{1-4r^2}}dr\tag5$

After further substitution $2r=u$ and using (2) the statement is proved:

$\frac{e^{2x}}{\pi}\int\limits_{-1}^{1} \dfrac{e^{2xu}}{\sqrt{1-u^2}}du=e^{2x}I_0(2x)\tag6$