Does this sum to $2^n-1\ $? $$N=\sum_{i=1}^{n} C^i_n = \sum_{i=1}^n\frac{n(n-1)\cdots(n-i+1)}{i!}$$
Does $N = 2^n-1$ hold ?
I mean, $C^i_n = \binom{n}{i}$. According to the binomial formula, if this summation sums from $i=0$ instead of $i=1$, then it's equal to $2^n$.
Because of this, does this sum to $2^n-1$?
 A: Yes. We know that
$$
\sum_{i=0}^n \binom{n}{i} = 2^n.
$$
Therefore
$$
\sum_{i=1}^n \binom{n}{i} = \sum_{i=1}^n \binom{n}{i} + \binom{n}{0} - \binom{n}{0} = \sum_{i=0}^n \binom{n}{i}-\binom{n}{0}
= 2^n-1.
$$
A: A more combinatorial way to look at this: 
We know that $\binom{n}{i}$ represents the number of ways to choose exactly $i$ elements from a collection of $n$ elements.
We sum this over $i=1,2,\ldots,n$; so, the result is the total number of ways to choose a set consisting of EITHER 1 element, 2 elements, 3 elements, ... up to $n$ elements, from a collection of $n$ elements.
In other words: it is the number of ways to choose a non-empty subset from a collection of $n$ elements.  There are $2^n$ ways to choose ANY subset (including the possibility of the empty subset); we can see this by noting that each element is either included or excluded, and we have to make $n$ such decisions.  And, there is only such sequence of choices that leads to the empty set.  So, the total number of non-empty subsets possible is just $2^n-1$.
A: We know that 
$$(a+b)^n=\sum_{i=0}^n \binom {n}{i}a^{n-i}b^i$$
with $a=b=1$, it becomes
$$2^n=\sum_{i=0}^n\binom {n}{i} $$
$$=\sum_{i=\color {red} {1}}^n\binom {n}{i}+\frac {n!}{0!(n-0)!}$$
$$=N+1$$
