Number of "rising/increasing sets"? I have the following problem to solve:

a) was pretty easy to show, but  I am struggling to count the sequences in b.
So far I noticed the obvious: $$|T_{i}|\geq i$$
Counting the sequences leads to a very ugly sum, I was wondering whether there's some clever combinatorial way to count them! Has a) to be used at some point?
I'd appreciate if someone could give me some help or a good hint.
Thanks in advance!
EDIT: $T_{i}$ are supposed to be subsets of $[n]:=\left \{ 1, ..., n \right \}$.
 A: HINT: Let $D_1=T_1$, for $i=2,\ldots,k$ let $D_i=T_i\setminus T_{i-1}$, and let $D_{k+1}=[n]\setminus T_k$. Notice that the sets $D_1,\ldots,D_{k+1}$ are pairwise disjoint, and all of them except possibly $D_{k+1}$ are non-empty. Moreover, knowing these sets is equivalent to knowing the sets $T_i$ for $i=1,\ldots,k$, so an equivalent problem is counting the pairwise disjoint families $\{D_1,\ldots,D_{k+1}\}$ of $k+1$ subsets of $[n]$ such that


*

*$\bigcup_{i=1}^{k+1}D_i=[n]$, and  

*$D_i\ne\varnothing$ for $i=1,\ldots,k$.


If we ignore the second restriction, there are $(k+1)^n$ such families: each $r\in[n]$ can be put into any of the $k+1$ sets $D_1,\ldots,D_{k+1}$.
Unfortunately, some of these distributions violate the second restriction, because they leave some $D_i$ empty; you can compensate for this with an inclusion-exclusion argument.
Alternatively, if you’re familiar with Stirling numbers of the second kind, you can note that what’s wanted here is almost $n\brace{k+1}$, the number of ways to partition $[n]$ into $k+1$ non-empty subsets, and make a small adjustment to cover the almost.
