$!n=\sum_{k=0}^n(-1)^{k-1}\frac{n!}{k!}$ (Proof by inclusion exclusion) This question is from here:
Let $S_i$ be the set of permutations of $n$ items which fix item $i$. Then the number of permutations in $k$ of the $S_i$ would be the permutations that fix $k$ items. There are $\binom{n}{k}$ ways to choose the $k$ items to fix, and $(n−k)!$ ways to arrange the other $n−k$ items. Thus, the number of permutations that fix at least $1$ item would be $$\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}(n-k)!=\sum_{k=1}^n(-1)^{k-1}\frac{n!}{k!}\tag{4}$$ Since there are $n!$ permutations in total, the number of permutations that don't fix any items is $$\begin{align}
\mathcal{D}(n)
&=n!-\sum_{k=1}^n(-1)^{k-1}\frac{n!}{k!}\\
&=\sum_{k=0}^n(-1)^k\frac{n!}{k!}\tag{5}
\end{align}$$
where $$\mathcal{D}(n)=!n$$
Now,I can't understand how the equation (4) is derived.
Please help me to understand this question. Thank you.
 A: Note that there are two things to (4). Firstly, the sentence leading up to this equation says that $\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}(n-k)!$ counts the number of permutations with at least one fixed point. This fact is obtained from the principle of inclusion-exclusion: for $k=1$ one counts for each of the $\binom n1$ points the $(n-1)!$ permutations fixing it, for $k=2$ one subtracts off for each of the $\binom n2$ pairs of points the $(n-2)!$ permutations fixing them both, and so forth. Then (secondly) (4) itself simplifies the expression by multiplying the $(n-k)!$ into the binomial coefficient: $\binom nk(n-k)!=\frac{n!}{k!}$.
By the way the whole derivation is rather roundabout, since a final step of inclusion-exclusion, subtracting off all permutations fixing at least one point from the total, is done "by hand" afterwards, while it could have been integrated into the inclusion-exclusion application itself. It could have said just

The number of permutations that fix $0$ items is
  $$
  \sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)!=\sum_{k=0}^n(-1)^k\frac{n!}{k!}.\tag{4a}
$$

This counts for $k=0$ all $\binom n0n!$ permutations, subtracts off ($k=1$) for each of the $\binom n1$ individual points the $(n-1)!$ permutations fixing it, for $k=2$ adds back etcetera.
