Let $F$ be a PRF, is the following $G(x)=F_{0…0}(x)∥F_{0…1}(x)|| ... ||F_{1…1}(x)$ a PRG? This exercise comes from Coursera's "Cryptography" course.


  Let $F$ be a PRF, is the following $G(x)=F_{0…0}(x)∥F_{0…1}(x)|| ... ||F_{1…1}(x)$ a PRG?


$F$ is a PRF with $n$ bit key, input, and output length. I know the answer is “NO it isn't!”, but I can't understand why. 

Is that because on any input $x$, the keys for every $F$ in the concatenation sequence are fixed, which implies that the $F$ functions are not randomly chosen over $2^{-n}$? Could this be a possible answer?
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.)