Let $F$ be a PRF, is the following $G(x)=F_{0…0}(x)∥F_{0…1}(x)|| ... ||F_{1…1}(x)$ a PRG?
Because the key-, input- and output-length of $F$ are equal, if the length of the seed $x$ is $n$, there are $2^n$ keys and so $G$ must compute $F$ $2^n$ times and output a string of length $n\cdot 2^n$. Hence, $G$ runs in time exponential in $n$, which means it can't be a pseudorandom generator because a pseudorandom generator is a deterministic polynomial-time algorithm.

(In addition, since a distinguisher runs in time polynomial in the length of $G(x)$, it runs in time exponential in the length of $x$, and brute force through all possible seeds is allowed.)