Generating fuctions: $\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}\binom{2n}{n}$ I'm trying to solve one of my combinatorics exercise but I struggle a bit.

Is the equality correct for all the $n\ge 0$? 
  $$\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}\binom{2n}{n}$$

First of all:$$\sum_{n=0}^\infty \left(2^{-2n}\binom{2n}{n}\right)x^n=\sum_{n=0}^\infty \left( \frac{x}{4}\right)^n\binom{2n}{n}=\frac{1}{\sqrt{1-x}}$$
Now
$$\sum_{n=0}^\infty \left(\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k \right)x^n=\sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{n=k}^\infty \binom{2k}{k}\binom{n}{k} x^n=[n-k=m]=\sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{m=0}^\infty \binom{2k}{k}\binom{m+k}{m} x^{m+k}=\sum_{k=0}^\infty \left(-\frac{x}{4}\right)^k\binom{2k}{k}\sum_{m=0}^\infty \binom{m+k}{m} x^{m}$$
and here I don't know what to do next. Can anyone help me? Thanks in advice!

You may consider that
$$ \frac{1}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sin^{2k}(\theta)\,d\theta \tag{A}$$
from which it follows that:
$$ \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sum_{k=0}^{n}\binom{n}{k}(-\sin^2\theta)^k =\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta\tag{B}$$
and the conclusion is straightforward through the substitution $\theta\mapsto\frac{\pi}{2}-\varphi$.

With generating functions, from 
$$ \sum_{k\geq 0}\binom{2k}{k}\frac{z^k}{4^k}=\frac{1}{\sqrt{1-z}} \tag{C}$$
by replacing $z$ with $\frac{x}{1+x}$, then by multiplying both sides by $\frac{1}{1+x}$, we get
$$ \sum_{k\geq 0}\binom{2k}{k}\frac{x^k}{4^k(1+x)^{k+1}} = \frac{1}{\sqrt{1-x}}\tag{D}$$
then by applying $[x^n]$ to both sides:
$$ \sum_{k\geq 0}\binom{2k}{k}\frac{(-1)^k}{4^k}\binom{n}{k} = \frac{1}{4^n}\binom{2n}{n}\tag{E}$$
where the RHS has been managed through $(C)$ and the LHS through stars and bars.
