Observe that with inclusion-exclusion we can count the number of
permutations having at least $k$ values where $q$ is followed by $q+1$
by choosing those $k$ from the range from $1$ to $n-1$ and fusing them
with the elements that follow immediately, forming at most $k$ blocks.
We may then permute these $n-k$ items any way we like.
This yields per inclusion-exclusion
$$\sum_{k=0}^{n-1} {n-1\choose k} (-1)^k (n-k)!
= (n-1)! \sum_{k=0}^{n-1} \frac{(-1)^k}{k!} (n-k)
\\ = (n-1)! \sum_{k=0}^{n} \frac{(-1)^k}{k!} (n-k)
= n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}
- (n-1)! \sum_{k=0}^{n} k \frac{(-1)^k}{k!}
\\ = D_n - (n-1)! \sum_{k=1}^{n} k \frac{(-1)^k}{k!}
= D_n - (n-1)! \sum_{k=1}^{n} \frac{(-1)^k}{(k-1)!}
\\ = D_n + (n-1)! \sum_{k=0}^{n-1} \frac{(-1)^k}{k!}
= D_n + D_{n-1}.$$
A block of length $m$ reduces the number of items being permuted by
$m$ and then adds one back in. On the other hand $m-1$ is the number
of adjacent consecutive items in the block and summing this over all
blocks yields $k.$