Does the (Stirling number of the second kind) equality ${2n\brace 2} = 2^{2n-1}-1$ hold? I filled in from the definition of a Stirling number of the second kind that the following holds.
$${2n\brace 2} = \frac{1}{2} \sum_{i=0}^{2} (-1)^i \binom{2}{i} (2-i)^{2n}$$
And I've visually confirmed in Desmos that the following equality 'appears' to hold.
$${2n\brace 2} = 2^{2n-1}-1$$
How do I prove that this equality does in fact hold?
 A: I guess the most straight-forward way is to expand the summation:
\begin{align}
{2n\brace 2} = &\frac{1}{2} \sum_{i=0}^{2} (-1)^i \binom{2}{i} (2-i)^{2n} = \\
& \frac{1}{2} \Bigg( \Big[(-1)^0 \binom{2}{0} (2-0)^{2n}\Big] + \Big[(-1)^1 \binom{2}{1} (2-1)^{2n}\Big] + \Big[(-1)^2 \binom{2}{2} (2-2)^{2n}\Big] \Bigg) = \\
&\frac{1}{2} \Bigg( \Big[1 \cdot 1 \cdot 2^{2n}\Big] + \Big[(-1) \cdot 2 \cdot 1^{2n}\Big] + \Big[1 \cdot 1 \cdot 0^{2n}\Big] \Bigg) = \\
&\frac{1}{2} \Bigg(  2^{2n} - 2 + 0 \Bigg) = \\
& \frac{2}{2} \Bigg(  2^{2n-1} - 1 \Bigg) = 2^{2n-1} - 1 
\end{align}
A: The definition of ${2n \brace 2}$ is dividing set $A= \{ a_1,...,a_{2n} \}$ into two sets which is not an empty set.
First, lets say that we are dividing set A to set B and set C.
For an integer $i$ such that $1 \le i \le 2n$,
$a_i \in B$ or $a_i \in C$, which means for every $i$, there are 2 cases.
Considering that B and C shouldn't be an empty set,
Dividing set A into set B and set C has $2^{2n}-2$ cases.
So, there are $2^{2n-1}-1$ cases.
