Recurrence relation for number of indecomposable permutations 
We say the permutation $\sigma = a_1,..., a_n \in S_n$ is indecomposable if
  ${a_1 , \dots ,a_j} \neq [j] ,\forall j<n$. Let $f(n)$ be the count of the indecomposable permutations in $S_n$. 
Show that,
$(a)$ $\sum_{j=1}^n f(j)(n-j)! = n!$ and $n\geq 1$
$(b)$ $(\sum_{n\geq1}f(n)t^n)(\sum_{n\geq0}n!t^n)=(\sum_{n\geq1}n!t^n)-1$

to i) let $A_i$ be the set of all perms $\sigma \in S_n$ with $\sigma[i]=[i]$ and $\sigma \in S_n$ with $\sigma[j]\neq[j]$ 
$\Rightarrow S_n = A_1 \cup A_2 \cup \dots \cup A_n$ 
why should this be $f(i)(n-i)$ ? 
to ii) no idea :-(

For the first part, we just double count the number of permutations of $[n]$. The right hand side is just the number of permutations, $n!$. On the left hand side, we group permutations based on the first index $i$ such that $a_1\dots a_i=[i]$. If the first such index is $i$, then there are $f(i)$ choices for the first $i$ letters, and then we just need an arbitrary permutation of $\{i+1,\dots,n\}$ to finish our permutation. So for a given $i$, there are $f(i)(n-i)!$ such permutations. Summing over all possible indices gives the desired result.
For the second part, read about multiplying power series (here is a link if you need: http://en.wikipedia.org/wiki/Power_series#Multiplication_and_division), and then use the first part.
