How likely is a AES KDF bypass? So I was wondering about some alternative attack scenario which I thought of when I started to understand KDF's . 
So lets say you have a cipher like AES and a strong KDF . And the legit user uses a 30 character strong Password :

30-char Pass ---> KDF ( slow stuff : argon2,scrypt ) ----> AES256

And normally the attacker would have to start at the beginning with brute force and then passing through the KDF which slowes him down very much and then finally checking if the final bit string is accepted by AES to decrypt the ciphertext .

But I the attacker knows that the legit user uses a very strong Pasword and a decent KDF , would it not be more efficient to just bypass the KDF and brute force the AES bits directly ( takes longer , but much less slowdown ) like : 

wordlist ---> Some random 256 hash ( sha256 ) -----> AES256   

With that method the attacker could brutefoce every possible AES key and the length would not even matter . But It would be very difficult because of all the possibilities .

How likely is such a Scenario ? And is it even worth doing so ?
Let's assume a password consisting of 15 random longish words from several books.  It's likely then that the password's entropy will exceed 256 bits. (This is actually very difficult to compute accurately as there are linguistic /social /physiological issues to human password creation, but let's skip over those).  That's more than that of an AES-256 key.  So in this case it seems that it's worthwhile attacking the AES key directly as it's got less entropy.

This would be a brute force attack directly against an AES-256 key.  The futility of this is well covered on this site.  I'll just add this link and say that you'd need more than all of the batteries on Earth for the energy demand. And a lot of time measured in age of Universe units.  It's impossible and really not worth doing.

It also opens up the question of what would KDFs be for if you simply bypassed them and attacked the cipher key directly?