How to show that $1 \over \sqrt{1 - 4x} $ generates $\sum_{n=0}^\infty \binom{2n}{n}x^n $

At first, consider the series $\sum_{n=0}^{\infty}C_{2n}^{n}x^{n}$, as $\frac{C_{2n+2}^{n+1}}{C_{2n}^{n}}=\frac{4(2n+1)}{2n+2}$, we can judge that the series convergent in $[-\frac{1}{4},\frac{1}{4})$.

Note $f(x)=\sum_{n=0}^{\infty}C_{2n}^{n}x^{n}=\sum_{n=0}^{\infty}\frac{(2n)!}{n!n!}x^{n}$, then $f(0)=1$, $$f^{\prime}(x) =\sum_{n=1}^{\infty}\frac{(2n)!}{n!(n-1)!}x^{n-1} =\sum_{n=0}^{\infty}\frac{(2n+2)!}{(n+1)!n!}x^{n} =\sum_{n=0}^{\infty}\frac{(4n+2)(2n)!}{n!n!}x^{n}$$ $$xf^{\prime}(x) =\sum_{n=1}^{\infty}\frac{(2n)!}{n!(n-1)!}x^{n} =\sum_{n=0}^{\infty}\frac{n(2n)!}{n!n!}x^{n}$$ $$(1-4x)f^{\prime}(x)=2\sum_{n=0}^{\infty}\frac{(2n)!}{n!n!}x^{n}=2f(x)$$ then solve the last ordinary differential equation, we get $f(x)=\frac{c}{\sqrt{1-4x}}$, as $f(0)=1$, so $f(x)=\frac{1}{\sqrt{1-4x}}$, i.e. $\sum_{n=0}^{\infty}C_{2n}^{n}x^{n}=\frac{1}{\sqrt{1-4x}}, x\in[-\frac{1}{4},\frac{1}{4})$.