How to prove the sum of combination is equal to $2^n - 1$
The binomial theorem states that $$\sum_{k=0}^n \binom{n}{k} x^k=(1+x)^n$$ Putting $x=1$ gives $$\sum_{k=0}^n \binom nk=2^n\\ \sum_{k=1}^n\binom nk=2^n-1$$
Taking all the rounds together (including the $0^{th}$), you have formed all combinations with any of the five letters taken or not, which you can do in $2\cdot2\cdot2\cdot2\cdot2$ ways. (Equivalently, all five bits numbers from $00000$ to $11111$, which is exactly $2^5$.)
For perfect rigor, one should show that there are no repetitions nor omissions.