Solve for (a-1) using a series - Logic Problem Solution

[Bounceback]

iiii(a-1)!iiiiiiiiiiiiii(a-1)!iiiiiiiiiiiii(a-1)!iiiiiiiiiiiiiiiiii(a-1)!
------------ii+ii------------ii+ii------------iiiiiii+ii------------ii=ii2^a-1
i(a-1)!*0!iiiiiiiiii(a-2)!*1!iiiiiiiii(a-3)!*2!iiii...iiiiiii0!*(a-1)!

In case you are having trouble reading that:
(a-1)!/((a-1!*0!)) + (a-1)!/((a-2)!*1!) + (a-1)!/((a-3)!*2!)... + (a-1)!/(0!*(a-1)!) = 2^a-1
Assuming a = positive integer

Essentially, I haven't had much experience with logs, and would be interested in a hint to making this a little more workable.

The next part incorporates part of a logic problem. If you haven't completed (click here), don't continue

Spoiler

Each term represents a number of combinations that occurs with a specific number of objects. For example, if you have 10 people and want to find out how many combinations there are of them going to a party (1,6,10), (1,3,5,9,10) etc; you can find out by finding how many different combinations of 1 person there are, how many combinations of two people, etc. Essentially, I developed a chart to represent it:
1 1 1 1
1 2 3 4
1 3 6 10
1 4 10 20

And a formula to find any term where x=row # and y=column #
iii(x+y-2)!
---------------
(x-1)!*(y-1)!
The series above represents a sum of the diagonal of the chart. Someone else also noticed that the same problem could be solved using binary, hence the fact that they equal each other.
I can go into more detail about how I developed this if requested

 

[Svein]

2^{a-1}=(1 + 1)^{a-1} =1^{a-1}+\frac{a-1}{1}1^{a-2}1^{1}+...