So I have to show that the number of partitions of $n \in \mathbb{N}$ is $2^{n-1} -1 $. ( the order is important ).
So here is my attempt. Please be strict. If you find any mistake or something that iritates you, let me know:
We rewrite $n \in \mathbb{N}$:
$n = 1 + \cdots + 1 $ .( we sum $1$ $n$-times). We want to find the number of compositions with at least two parts, so we consider "$+$". For each "$+$" between the summands we have to decide if the summands are a "new" summand or not. We get $2^{n-1}$ decisions. But we don't count the solution $n = n$ , so we have $2^{n-1} -1$.
What do you say? Remark: Please focus on my work. Thank you in advance.
The wikipedia page https://en.wikipedia.org/wiki/Composition_(combinatorics) confirms both your answer and your argument.