There are two problems with your attempt.
First, you've used n+1 instead of i+1, so you're going to return something like [5, 5, 5, 5] instead of [1, 2, 3, 4].
Second, you can't for-loop over a number like n, you need to loop over some kind of sequence, like range(n).
So:
def naturalNumbers(n):
return [i+1 for i in range(n)]
But if you already have the range function, you don't need this at all; you can just return range(1, n+1), as arshaji showed.
So, how would you build this yourself? You don't have a sequence to loop over, so instead of for, you have to build it yourself with while:
def naturalNumbers(n):
results = []
i = 1
while i <= n:
results.append(i)
i += 1
return results
Of course in real-life code, you should always use for with a range, instead of doing things manually. In fact, even for this exercise, it might be better to write your own range function first, just to use it for naturalNumbers. (It's already pretty close.)
There is one more option, if you want to get clever.
If you have a list, you can slice it. For example, the first 5 elements of my_list are my_list[:5]. So, if you had an infinitely-long list starting with 1, that would be easy. Unfortunately, you can't have an infinitely-long list… but you can have an iterator that simulates one very easily, either by using count or by writing your own 2-liner equivalent. And, while you can't slice an iterator, you can do the equivalent with islice. So:
from itertools import count, islice
def naturalNumbers(n):
return list(islice(count(1), n))