Let us say $n = 5$.
The integers $1, 2, 3, 4$ are below it and $1, 2, 3$ are not divisible by 4. There are 4 numbers below $5$ and $3/4$ are not divisible by 5.
Now consider 9. The integers below it are $1, 2, 3, 4, 5, 6, 7, 8$ and again we see that $1, 2, 3, 5, 6, 7$ are not divisible by 4, which makes $6/8 = 3/4$ of them.
Now imagine a really large $n $: you have the integers $1, 2, 3, 4, 5, \cdots, n-1$. Now you group those number 4 by 4, like so: $(1, 2, 3, 4), (5, \cdots), \cdots, (\cdots, n-1) $. How many groups are there. Well, there are $n/4$ groups. Almost. For some $n $ it might not be exactly the case right? For small $n $, say 6, we could only form one complete group. Some $n $ do not allow only full groups, leaving 1, 2 or 3 numbers without a group. But for now let us assume there are $n/4$ complete groups.
Then 3 out of every 4 numbers of each group would not be divisible by 4 and hence $3\times n/4$ numbers would not be divisible by four. But $3\times n/4$ corresponds exactly to $3/4$ of the numbers.
If $n $ gets really large, $3n/4$ gets really large and you don't even bother with missing by 1 or 2 or 3 when $n $ is such that it doesn't allow you to only form full groups.
What you did for 4, you can do for any $k $, thus showing that for $n $ big enough, $\frac{k-1}{k} $ of the numbers are not divisible by $k $ (which is the same as saying that $1/k $ of them are.