I created my own drop' function like this:

drop' :: Int -> [a] -> [a]

drop' 0 xs = xs

drop' n [] = []

drop' n (x:xs) = drop' (n - 1) xs

But in the Haskell book I am reading it is written like this:

drop' :: Int -> [a] -> [a]

drop' 0 xs = xs

drop' (n + 1) [] = []

drop' (n + 1) (x:xs) = drop' n xs

My version seems to work but I am concerned about my third line (drop' n [] = []). Is that a problem?

Why does this author use n + 1 as a parameter as opposed to n - 1 like I do in body?

Or does it make no difference?