Why is $\sum_{j=1}^{n-1}[\Pi_{k=1}^{j}[(n-k)]]=2^n$?
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.