How many wrong envelopes?

Let $S_n$ is the total cases for total $n$ letters in $n$ envelopes, and $F_n$ is the cases for all of them in wrong envelope.

Obviously $S_n = n!$ , and $F_1 = 0$ and $F_n$ will be total cases minus exact 1~n letters in correct envelope respectively but all others are in wrong (the case for all letters in correct envelope always is 1). i.e. $\\ F_n = S_n - \sum\limits_{i=1}^{n-1} (\tbinom{i}{n} \cdot F_{n-i}) - 1$