In CLRS book, in the road cutting example there is a recursion formula $$    1+\sum_{j=0}^{n-1}T(j) $$and it can be proved that the sum is $$        2^n $$by simple induction. In 3-rd edition it is a formula 15.3 on a page 364.As I can understand from the recursion-tree aproach, the recursion tree which can be built for the said problem is $$    1->(n-1)->(n-2)|each.node->(n-3)|each.node->...$$and so on, which can be summed as $$    \sum_{j=1}^{n-1}[\Pi_{k=1}^{j}[(n-k)]]$$How can I come up with the suggestion of $2^n$ from this recursion tree (or why this tree is wrong and what can be the right approach to suggesting $2^n$ running time)?
Why is $\sum_{j=1}^{n-1}[\Pi_{k=1}^{j}[(n-k)]]=2^n$?
recurrence relation
Minor nit: your first displayed expression should be$$T(n) = \begin{cases}  1 & \text{if $n=0$,}\\   1 + \sum_{j=0}^{n-1}T(j) & \text{otherwise}\end{cases}$$You can indeed solve this by looking at the recurrence tree, but you don't wind up with your sum of products expression (try it for $n=4$ to see that). For this problem there's an easier way, though:Hint: What is $T(n+1)-T(n)$?You can take a look at our reference question, on recurrence relations, as well.