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 $2^n-1$?

We know that 
$$(a+b)^n=\sum_{i=0}^n \binom {n}{i}a^{n-i}b^i$$
with $a=b=1$, it becomes
$$2^n=\sum_{i=0}^n\binom {n}{i} $$
$$=\sum_{i=\color {red} {1}}^n\binom {n}{i}+\frac {n!}{0!(n-0)!}$$
$$=N+1$$
