Summation with running indices that have running indices I would like to be able to simplify an expression such as
$$\prod_{l=1}^{i} \sum_{j_l=j_{l-1}+1}^{m-i+l-1} f(j_l),$$
with $j_o:=0$.
For instance for $i=3,m=4$ the result is $f(1)f(2)f(3)$
whereas for $i=2,m=4$ the results is $f(1)f(2) + f(1)f(3) + f(2)f(3)$ and so on. In particular it would be great to get rid of the index sub_index, such that I can put my summations to something like maple for different f(x).
Many thanks for any hint!
 A: 
Regrettably this is not possible in the general case.
Let's have a look at the special case $i=3$. As already indicated by the examples in the question we can rearrange these sums to get
  \begin{align*}
\prod_{l=1}^3\sum_{j_l=j_{l-1}+1}^{m+l-4}f\left(j_l\right)
&=\sum_{j_1=1}^{m-3}f\left(j_1\right)\sum_{j_2=j_1+1}^{m-2}f\left(j_2\right)\sum_{j_3=j_2+1}^{m-1}f\left(j_3\right)\\
&=\sum_{1\leq j_1<j_2<j_3\leq m-1}f\left(j_1\right)f\left(j_2\right)f\left(j_3\right)\\
&=\sum_{1\leq j_1<j_2<j_3\leq m-1}\prod_{l=1}^3 f\left(j_l\right)
\end{align*}
In general we obtain
  \begin{align*}
\prod_{l=1}^i\sum_{j_l=j_{l-1}+1}^{m-i+l-1}f\left(j_l\right)=\sum_{1\leq j_1<j_2<\ldots<j_i\leq m-1}\prod_{l=1}^i f\left(j_l\right)
\end{align*}
  which exchanges the position of summation and product sign but does not simplify the index notation.

It depends on the properties of $f$ if e.g. due to symmetry the region of summation can be simplified.
Hint: You may want to have a look at Combinatorial Identities, Vol. 1 by H.W. Gould which provides many formulas with finite sums and products which could be useful.
