Why is $\sum_{j=1}^{n-1}[\Pi_{k=1}^{j}[(n-k)]]=2^n$? 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 
$$
    1-&gt;(n-1)-&gt;(n-2)|each.node-&gt;(n-3)|each.node-&gt;...
$$

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)?
Minor nit: your first displayed expression should be
$$
T(n) = \begin{cases}
  1 &amp; \text{if $n=0$,}\\ 
  1 + \sum_{j=0}^{n-1}T(j) &amp; \text{otherwise}
\end{cases}
$$
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.