Probability of drawing numbers without any number being in its sequential position

Based on a joke going around in Video Game randomizers, where items that are appearing to be “not randomized” are thought to be broken, I was interested in the probabilities related to this.

To simplify, we have n items, which are randomized to appear in n positions.

This can be considered to be similar to be having a bag of n tokens, numbered 1 to n, and we draw them sequentially, so that in a bag of 3, a draw of 3-2-1 is counted as a draw in sequential order, as 2 is drawn in position 2.

Now, the probability of drawing them all in order is simple, $frac{1}{n!}$.
But what is the probability of none of them being in order? (Because, obviously, none of them being in their original order is “truly” random…..) Based on a question and comment here it seems that Rencontres numbers fit my empirical test at low numbers, where

n No sequential draw occurrences Probability
3 2 2/6 = 1/3
4 9 9/24 = 3/8
5 44 44/120 = 11/30

So am I correct in saying that we can use the formula for Rencontres numbers here, which is

$$D_{n,0} = leftlceilfrac{n!}{e}rightrfloor$$

?

Which interestingly enough, without looking too hard on expanding that formula, seems to tend to around 0.3678 as n gets bigger.