The exponential generating function for the central binomial coefficients I am having difficulty proving the following result
\begin{align}
\sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{n}}{n!} = e^{2x} \ I_{0} (2x) \text{,}
\end{align}
where 
\begin{align}
I_{0}(y) = \sum_{n=0}^{\infty} \frac{\Big( \frac{y}{2} \Big)^{2n}}{n! \ n!}
\end{align}
is a modified Bessel function of the first kind with zero order. 
I approached it using
\begin{align}
e^{2x} \ I_{0} (2x) = \bigg( \sum_{n=0}^{\infty} \frac{(2x)^{n}}{n!}\bigg) \ \bigg( \sum_{k=0}^{\infty} \frac{ x^{2k}}{k! \ k!} \bigg) \text{,}
\end{align}
but I have had no luck.
 A: 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.
A: 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: 
$\frac{e^{2x}}{\pi}\int\limits_{-1}^{1} \dfrac{e^{2xu}}{\sqrt{1-u^2}}du=e^{2x}I_0(2x)\tag6$
