Why is $FOLLOW$ not necessary for $LL(1)$ grammars with no $\epsilon$ transitions? I'm aware of how $FIRST$ and $FOLLOW$ sets are used to construct a parsing table for $LL(1)$ grammars.

However, I've encountered this statement from my notes:


  With $\epsilon$ productions in the grammar, we may have to look
  beyond the current non-terminal to what can come after it


In my opinion, this suggests that $FOLLOW$ is not necessary for $LL(1)$ grammars that have no $\epsilon$ transition. Am I wrong? And if I'm not, why is this the case?

Thanks
That's correct for $LL(1)$: if there are no $\epsilon$ productions, the $LL$ parser-generation algorithm will never consult $FOLLOW$, because it only does that if it finds $\epsilon$ in the $FIRST$ set for the first non-terminal in the right-hand side of a production. (So it might not need the $FOLLOW$ sets even if there are some $\epsilon$ productions, provided that none of those productions occur at the beginning of a right-hand side.) 

The observation doesn't generalise well to other values of $k$. You'll need $FOLLOW_k(\alpha)$ if any production can derive a string whose length is less than $k$.