How many ways to divide $n$ different pieces of chocolate in two non empty groups? After the example I think that the order of the groups doesn't matter so ${(A),(B,C)}$ and $(B,C),(A)$ counted as $1$.
Suppose we split $5$ chocolates into a group of size $1$ and a group of size of $4$. So there are $5$ different ways.
Next split the numbers into a group of size $2$ and a group of size $3$. Using the binomial coefficient I get $10$ ways to do this.
So all together for $n=5$ I found $15$ ways.
I've started to calculate the $n$ for "little" numbers just like, $1,2,3,4,5$ and I got the assumption that the formula is $2^{n-1}-1$.
Is that true? And how can I show that my assumption is correct?

One way is to consider putting some/all/none of the chocolates on the left and the rest on the right.  Each chocolate can go either way, so there are $2^n$ possibilities.
But $2$ of these possibilities leave one side empty.  Ignoring these, there are $2^n-2$ possibilities.
Each pattern has a distinct reflection, swapping all those on the left with all those on the right.  If you regard reflections as equivalent patterns, then there are $\frac{2^n-2}{2} = 2^{n-1}-1$ possibilities. 
