It's rather common knowledge to know that if we want to rearrange $N$ distinct items' positions such that none of them appears in its original position, then it is a derangement problem. So if we rearrange 7 distinct items' position, then the probability that none of them appears in their original position is $ \frac{!7}{7!} = \frac{1854}{5040}$.
Here's a follow-up question:
Suppose this time we have $N$ distinct items arranged from left to right. We start from the left to right and count how many items are not in their original position, and we stop counting once we have found an item in the original position. $$$$ To illustrate this, consider a permutation of the 5 items $(A,B,C,D,E)$ to be $(B,A,C,E,D)$. This tells us that we've only found 2 wrong positions, instead of 4 (because we stopped searching after discovering that $C$ is in its original position). $$$$ Let $\mathsf B_N$ denote the discrete probability distribution for the number of these "partial derangements" after permutating $N$ items. It's easy to construct $\mathsf B_N$ for small $N$'s: $$ \mathsf B_2 = \begin{cases} 0, \quad \mathsf{Pr} = 1/2 \\ 2, \quad \mathsf{Pr} = 1/2 \end{cases} \quad\quad, \quad\quad \mathsf B_3 = \begin{cases} 0, \quad \mathsf{Pr} = 2/6 \\ 1, \quad \mathsf{Pr} = 1/6 \\ 2, \quad \mathsf{Pr} = 1/6 \\ 3, \quad \mathsf{Pr} = 2/6 \\ \end{cases} $$ Here's the directive: For all integers $n\geqslant 2$, prove the probability mass function $$ \mathsf{Pr}[\mathsf B_n = k] = \left \{ \begin{array} {l l } \left [ \sum \limits_{j=0}^k (-1)^j \cdot \binom kj \cdot (n-j-1)! \right ]/n! \,,& k = 0,1,2,\ldots , n-2 \\ !(n-1) /n! \, , & k=n-1 \\ !n /n! \,, & k=n \end{array} \right . $$
Here are a few obvious observations:
$\mathsf{Pr}[\mathsf B_n = 0 ] $ simplifies to $ \frac1n$, which is true because this means that the first item (after shuffled) is already in the original position.
$\mathsf{Pr}[\mathsf B_n = n ] $ is simply a derangement problem.
$\mathsf{Pr}[\mathsf B_n = n-1 ] $ is simply a derangement problem as well but only the rightmost item is in the correct position.
How did I come up with this? I was playing around with slightly larger $N$'s, and with some Monte Carlo, I'm able to construct the empirical formulas for $\mathsf B_N$. Then I noticed that the numerators of the probability fractions for each of these specific $N$'s follow a nice sequence. E.g.:
$$ \mathsf B_8 = \begin{cases} 0, \quad \mathsf{Pr} = 7!/8!\\ 1, \quad \mathsf{Pr} = (7! - 6!)/8! \\ 2, \quad \mathsf{Pr} =(7! - 2\cdot 6! + 5!)/8! \\ 3, \quad \mathsf{Pr} = (7! - 3\cdot 6! + 3\cdot 5! - 4!)/8!\\ 4, \quad \mathsf{Pr} = (7! - 4\cdot 6! + 6\cdot 5! - 4\cdot 4! + 3!)/8! \\ 5, \quad \mathsf{Pr} = (7! - 5\cdot 6! + 10\cdot 5! - 10\cdot 4! + 5 \cdot 3! - 2!)/8! \\ 6, \quad \mathsf{Pr} = (7! - 6\cdot 6! + 15\cdot 5! - 20\cdot 4! + 15 \cdot 3! - 6\cdot 2! + 1!)/8! \\ 7, \quad \mathsf{Pr} = (!7)/8! \\ 8, \quad \mathsf{Pr} = \, !8/8! \\ \end{cases} $$
Essentially the same inclusion-exclusion argument that yields the yields the result
$$d_n=n!\sum_{i=0}^n\frac{(-1)^i}{i!}$$
also yields the result
$$\mathsf{Pr}[\mathsf{B}_n=k]=\frac1{n!}\sum_{i=0}^k(-1)^i\binom{k}i(n-i-1)!$$
for $k=0,\ldots,n-2$. For $i\in[k]$ let $S_i$ be the set of permutations of $[n]$ that fix $i$ and $k+1$. Then
$$\begin{align*} \left|\bigcup_{i=1}^kS_i\right|&=\sum_{\varnothing\ne I\subseteq[k]}(-1)^{|I|+1}(n-1-|I|)\\ &=\sum_{i=1}^k(-1)^{i+1}\binom{k}i(n-i-1)!\,, \end{align*}$$
So the number of permutations of $[n]$ for which $k+1$ is the smallest fixed point is
$$\begin{align*} &(n-1)!-\sum_{i=1}^k(-1)^{i+1}\binom{k}i(n-i-1)!\\ &\qquad=\sum_{i=0}^k(-1)^i\binom{k}i(n-1-i)!\,, \end{align*}$$
as desired.